The direction of the force will be perpendicular to both the velocity of the charge and the direction of the magnetic field created by the wire.
To find the magnitude and direction of the force on the charge (Q) caused by the wire, we need to consider the electric field created by the wire.
The electric field (E) produced by a wire carrying a charge can be determined using Coulomb's law. The electric field is given by the equation:
E = k * (Q / r²),
where k is the electrostatic constant (8.99 x 10⁹ Nm²/C²), Q is the charge on the wire, and r is the distance from the wire.
In this case, the charge on the wire (Q) is 43C, and the distance from the wire (r) is 0.2m. Substituting these values into the equation, we have:
E = (8.99 x 10⁹ Nm²/C²) * (43C / (0.2m)²).
Next, we can calculate the force (F) experienced by the charge (Q) using the equation:
F = Q * E.
Plugging in the value for the charge (Q) and the electric field (E), we get:
F = 43C * E.
Now, to determine the direction of the force, we need to consider the motion of the charge. Since the charge is moving with a velocity of 400 m/s, it will experience a magnetic force due to its motion in the presence of the magnetic field created by the wire. The direction of this force can be determined using the right-hand rule.
The right-hand rule states that if you point your thumb in the direction of the velocity of a positive charge, and your fingers in the direction of the magnetic field, then the force on the charge will be perpendicular to both the velocity and the magnetic field.
Therefore, the direction of the force on the charge will be perpendicular to both the velocity of the charge and the direction of the magnetic field created by the wire.
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Charge on String in Electric Field In this problem you must determine the charge on a pith ball that is suspended in a charged capacitor. You will be given the mass of the pith ball, the angle that the string makes with the vertical and the gravitational field of the planet on which this system is located. You will also be told the potential difference between the plates of the capacitor and the distance between the plates of the capacitor. You can ignore edge effects of the capacitor. Finally, you must find the tension in the string holding the pith ball. When you are ready to start this activity, click on the begin button. Begin 1203 Awe Charge on String in Electric Field 1 1 1 1173 V Enter Answers Show Question 1 Charge on String in Electric Field The gravitational field of this planet is 6.7 N/kg The mass of the ball is 393.0 mg. The potential differnece between the plates of the capacitor is 1173 V. The distance between the plates of the capacitor is 52.0 mm. The string makes an angle of 37.82° with the vertical. Determine the tension in the string. Determine the charge on the ball. When you are ready test your answers, hit the 'Enter Answers' Button 1173 V Enter Answers Hide Question Charge on String in Electric Field I Enter Your Answers Below Don't Enter Units Your Name: Charge (nC): Tension (mN): Submit 1173 V Hide Answers Show Question
To determine the tension in the string and the charge on a pith ball suspended in a charged capacitor.
To find the tension in the string, we need to consider the forces acting on the pith ball. There are two forces: the gravitational force and the electrostatic force.
Gravitational Force:
The gravitational force acting on the pith ball can be calculated using the mass of the ball (393.0 mg) and the gravitational field of the planet (6.7 N/kg). We can use the equation F_gravity = m * g, where m is the mass and g is the gravitational field.
F_gravity = (393.0 mg) * (6.7 N/kg)
Electrostatic Force:
The electrostatic force experienced by the pith ball is given by Coulomb's law, which states that the electrostatic force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.
Since the pith ball is suspended in a charged capacitor, the electrostatic force is balanced by the tension in the string. Therefore, the tension in the string is equal to the electrostatic force.
To find the electrostatic force, we need to determine the charge on the pith ball. This can be done by considering the potential difference between the plates of the capacitor and the distance between the plates.
Using the equation V = Ed, where V is the potential difference, E is the electric field, and d is the distance between the plates, we can find the electric field E.
E = V / d
Once we have the electric field, we can calculate the electrostatic force using the equation F_electrostatic = q * E, where q is the charge on the pith ball.
Tension in the String:
Since the tension in the string balances the gravitational force and the electrostatic force, we can equate these forces:
F_gravity = F_electrostatic
From this equation, we can solve for the tension in the string.
Charge on the Ball:
To find the charge on the pith ball, we can rearrange the equation for the electrostatic force:
F_electrostatic = q * E
We already know the electric field E, and we can substitute the calculated tension in the string as the electrostatic force to solve for the charge q.
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Is it possible that
the resitivity of gold is not 2.44x10^8?
It is possible for the resistivity of gold to deviate from this value under certain conditions or due to impurities.
The resistivity of gold is a physical property that can be measured experimentally. The standard value for the resistivity of gold at room temperature is approximately 2.44 x 10^-8 ohm-meters. However, it is possible for the resistivity of gold to deviate from this value due to various factors such as impurities, temperature, pressure, and strain.
For example, the resistivity of gold can increase with increasing temperature, as the thermal energy causes the gold atoms to vibrate more and impede the flow of electrons. Similarly, the resistivity of gold can also increase under high pressure, as the movement of electrons is restricted by the compression of the gold lattice. Furthermore, the presence of impurities or defects in the gold lattice can also affect its resistivity.
Therefore, while the standard value for the resistivity of gold is 2.44 x 10^-8 ohm-meters, it is possible for the resistivity of gold to deviate from this value under certain conditions or due to impurities.
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(K=2, C=1) \) Define the following terms in your own words: (a) Standing Wave : (b) node:"
A standing wave is a wave pattern that is created by the superposition of two identical waves traveling in opposite directions. A node is a point in a standing wave pattern where the amplitude is zero
(a) Standing Wave: A standing wave is a wave pattern that is created by the superposition of two identical waves traveling in opposite directions. The superposition of these waves produces a pattern of the wave that does not appear to move. Instead, it vibrates in place and maintains its position while oscillating between its minimum and maximum amplitudes. It is important to note that in a standing wave, the energy is not transmitted across the medium, as the waves simply oscillate in place.
(b) Node: A node is a point in a standing wave pattern where the amplitude is zero. It is the point in the wave where the two opposing waves cross and cancel each other out, causing no displacement to occur. In other words, a node is the point of minimum energy and maximum stability in a standing wave. Nodes can occur at regular intervals along the wave pattern, depending on the frequency of the wave. For example, a wave with a frequency of 150 Hz would have nodes occurring at every half-wavelength (which is equivalent to a distance of 0.85 meters, assuming a speed of sound of 340 m/s).
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A girl is wearing a cowgirl boot to a square dance. Estimate the pressure exerted on the dance floor by each heal
if the pointed heel has an area = 0.23 cm°2 and the cowgirl has a mass of 58.2-kg.
The pressure exerted on the dance floor by each heel of the cowgirl's boot is approximately 25,224 Pascal (Pa).
To estimate the pressure exerted on the dance floor by each heel, we can use the formula:
Pressure = Force / Area
We are given:
Area = 0.23 cm² (converted to square meters, 1 cm² = 0.0001 m²),
Mass = 58.2 kg (mass of the cowgirl).
We need to calculate the force exerted by the cowgirl's heel. The force can be determined using Newton's second law:
Force = mass * acceleration
Since the cowgirl is standing still on the dance floor, the acceleration is zero, and therefore the net force acting on her is zero. However, to calculate the pressure exerted on the dance floor, we need to consider the normal force exerted by the cowgirl on the floor.
The normal force is equal in magnitude and opposite in direction to the force exerted by the cowgirl's heel on the floor. Therefore, we can use the weight of the cowgirl as the force exerted by each heel.
Weight = mass * gravitational acceleration
Gravitational acceleration is approximately 9.8 m/s².
Weight = 58.2 kg * 9.8 m/s²
Now we can calculate the pressure:
Pressure = Force / Area
= Weight / Area
Substituting the values:
Pressure = (58.2 kg * 9.8 m/s²) / 0.23 cm²
First, let's convert the area to square meters:
Area = 0.23 cm² * 0.0001 m²/cm²
Pressure = (58.2 kg * 9.8 m/s²) / (0.23 cm² * 0.0001 m²/cm²)
Calculating:
Pressure ≈ 25,224 Pa
Therefore, the pressure exerted on the dance floor by each heel of the cowgirl's boot is approximately 25,224 Pascal (Pa).
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What quantum numbers are needed to give a complete
description of the quantum state of an electron in an atom?
(b) List the value of each of the quantum numbers mentioned in (a) for each of the
electrons in a neutral strontium atom (Z = 38) in its ground state.
The values of the quantum numbers for each electron in a neutral strontium atom (Z = 38) in its ground state are determined by the electron configuration and the rules governing the filling of electron orbitals.
To give a complete description of the quantum state of an electron in an atom, the following quantum numbers are needed:
Principal Quantum Number (n): It determines the energy level and average distance of the electron from the nucleus. Its values are positive integers starting from 1.Angular Momentum Quantum Number (ℓ): It determines the shape of the orbital and the magnitude of the orbital angular momentum. Its values range from 0 to (n-1).Magnetic Quantum Number (mℓ): It determines the orientation of the orbital in space. Its values range from -ℓ to ℓ, including 0.Spin Quantum Number (ms): It describes the intrinsic angular momentum or spin of the electron. It can have two possible values: +1/2 (spin-up) or -1/2 (spin-down).Now, let's list the values of each quantum number for the electrons in a neutral strontium atom (Z = 38) in its ground state:
The electronic configuration of strontium (Sr) in its ground state is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s²
1. For the 1s² electrons:
- n = 1
- ℓ = 0
- mℓ = 0
- ms = +1/2 (two electrons with opposite spins)
2. For the 2s² electrons:
- n = 2
- ℓ = 0
- mℓ = 0
- ms = +1/2 (two electrons with opposite spins)
3. For the 2p⁶ electrons:
- n = 2
- ℓ = 1
- mℓ = -1, 0, +1
- ms = +1/2 (six electrons with opposite spins)
4. For the 3s² electrons:
- n = 3
- ℓ = 0
- mℓ = 0
- ms = +1/2 (two electrons with opposite spins)
5. For the 3p⁶ electrons:
- n = 3
- ℓ = 1
- mℓ = -1, 0, +1
- ms = +1/2 (six electrons with opposite spins)
6. For the 4s² electrons:
- n = 4
- ℓ = 0
- mℓ = 0
- ms = +1/2 (two electrons with opposite spins)
7. For the 3d¹⁰ electrons:
- n = 3
- ℓ = 2
- mℓ = -2, -1, 0, +1, +2
- ms = +1/2 (ten electrons with opposite spins)
8. For the 4p⁶ electrons:
- n = 4
- ℓ = 1
- mℓ = -1, 0, +1
- ms = +1/2 (six electrons with opposite spins)
9. For the 5s² electrons:
- n = 5
- ℓ = 0
- mℓ = 0
- ms = +1/2 (two electrons with opposite spins)
So, in a neutral strontium atom (Z = 38) in its ground state, there are a total of 38 electrons.
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For a particular RL.C parallel circuit connected to an AC voltage source, the capacitive reactance is 13.96 S, the inductive
reactance is 24.3 S2, and the maximum voltage across the 75.9-S resistor is 14.5 V. What is the total current in the circuit?
Using the impedance triangle method, the total impedance of a parallel RL.C circuit was calculated to be 77.67 Ω. The maximum current in the circuit was calculated to be approximately 0.1865 A given the value of the maximum voltage across the resistor.
To solve this problem, we can use the impedance triangle method for a parallel RL.C circuit.
The total impedance Z of the circuit can be calculated as follows:
Z = sqrt((R-XC)^2 + XL^2)
Substituting the given values, we get:
Z = sqrt((75.9 - 13.96)^2 + 24.3^2)
Z = 77.67 Ω
The maximum current I in the circuit can be calculated using Ohm's law:
I = V_max / Z
Substituting the given values, we get:
I = 14.5 V / 77.67 Ω
I = 0.1865 A
Therefore, the total current in the parallel RL.C circuit is approximately 0.1865 A.
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4. A circular coil of wire with 20 turns and a radius of 40.0 cm is laying flat on a horizontal tabletop. There is a uniform magnetic field extending over the entire table with a magnitude of 5.00 T and directed to the north and downward, making an angle of 25.8° with the horizontal. What is the magnitude of the magnetic flux through the coil? 5. An 8-turn coil has square loops measuring 0.200 m along a side and a resistance of 3.00 Q2. It is placed in a magnetic field that makes an angle of 40.0° with the plane of each loop. The magnitude of this field varies with time according to B = 1.50t³, where t is measured in seconds and B in teslas. What is the induced current in the coil at t = 2.00 s?
The magnitude of the magnetic flux through the circular coil is approximately 2.275 T·m² when a uniform magnetic field of 5.00 T makes an angle of 25.8° with the normal to the coil's plane.
1. To find the magnitude of the magnetic flux through the circular coil, we can use the formula Φ = B * A * cos(θ), where Φ is the magnetic flux, B is the magnetic field, A is the area of the coil, and θ is the angle between the magnetic field and the normal to the coil.
2. First, we need to calculate the area of the coil. Since it is a circular coil, the area can be calculated as A = π * r^2, where r is the radius of the coil.
3. Substituting the given values, we find A = π * (0.4 m)^2 = 0.16π m².
4. Next, we calculate the cosine of the angle between the magnetic field and the normal to the coil.
Using the given angle of 25.8°, cos(θ) = cos(25.8°) = 0.902.
5. Now, we can calculate the magnetic flux using the formula: Φ = B * A * cos(θ).
Substituting the given values,
we have Φ = (5.00 T) * (0.16π m²) * (0.902) ≈ 2.275 T·m².
Therefore, the magnitude of the magnetic flux through the coil is approximately 2.275 T·m².
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5. Build a 2-input AND gate using CMOS.
To build a 2-input AND gate using CMOS (Complementary Metal-Oxide-Semiconductor) technology, we can use a combination of n-channel and p-channel MOSFETs (Metal-Oxide-Semiconductor Field-Effect Transistors).
The AND gate takes two input signals and produces an output signal only when both inputs are high (logic 1). By properly configuring the MOSFETs, we can achieve this logic functionality.
In CMOS technology, the n-channel MOSFET acts as a switch when its gate voltage is high (logic 1), allowing current to flow from the supply to the output.
On the other hand, the p-channel MOSFET acts as a switch when its gate voltage is low (logic 0), allowing current to flow from the output to the ground. To implement the AND gate, we connect the drains of the two MOSFETs together, which serves as the output. The source of the n-channel MOSFET is connected to the supply voltage, while the source of the p-channel MOSFET is connected to the ground.
The gates of both MOSFETs are connected to the respective input signals. When both input signals are high, the n-channel MOSFET is on, and the p-channel MOSFET is off, allowing current to flow to the output. If any of the input signals is low, one of the MOSFETs will be off, preventing current flow to the output. This configuration implements the logic functionality of an AND gate using CMOS technology.
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The specific heat of ice is 0.5 cal/g. "C and the latent heat of fusion is 80 cal/g. How much heat is absorbed by a 6 g of ice at -7°C to turn into 6 g of water at 0°C?
What is 24°C in Fahrenhiet?
What is 11F in Kelvin?
The heat absorbed by the 6g of ice to turn into water is 501 calories.
24°C is equal to 75.2°F and 11°F is approximately equal to 262.59 Kelvin.
To calculate the heat absorbed by the ice to turn into water, we need to consider two parts: heating the ice to its melting point and then melting the ice into water.
1. Heating the ice to its melting point:
The formula to calculate the heat absorbed for a temperature change is:
Q = m * c * ΔT
Where:
Q is the heat absorbed,
m is the mass of the ice (6 g),
c is the specific heat of ice (0.5 cal/g°C),
ΔT is the change in temperature (0°C - (-7°C) = 7°C).
Q1 = 6 g * 0.5 cal/g°C * 7°C
Q1 = 21 cal
2. Melting the ice into water:
The heat absorbed during the phase change (melting) is given by:
Q2 = m * L
Where:
Q2 is the heat absorbed during melting,
m is the mass of the ice (6 g),
L is the latent heat of fusion (80 cal/g).
Q2 = 6 g * 80 cal/g
Q2 = 480 cal
Total heat absorbed = Q1 + Q2
Total heat absorbed = 21 cal + 480 cal
Total heat absorbed = 501 cal
Therefore, 501 calories of heat is absorbed by the 6 g of ice to turn into 6 g of water.
To convert from Celsius to Fahrenheit, we use the formula:
°F = (°C * 9/5) + 32
24°C in Fahrenheit:
24°F = (24 * 9/5) + 32
24°F = 43.2 + 32
24°F = 75.2
Therefore, 24°C is equal to 75.2°F.
To convert from Fahrenheit to Kelvin, we use the formula:
K = (°F + 459.67) * 5/9
11°F in Kelvin:
K = (11 + 459.67) * 5/9
K = 472.67 * 5/9
K ≈ 262.59
Therefore, 11°F is approximately equal to 262.59 Kelvin.
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13 Select the correct answer. Which missing item would complete this alpha decay reaction? + He 257 100 Fm → OA. 29C1 253 98 B. 255 C. 253 D. 22th 904 O E. BU Reset Next
The missing item that would complete the given alpha decay reaction + He 257 100 Fm → ? is option C. 253.
In an alpha decay reaction, an alpha particle (consisting of two protons and two neutrons) is emitted from the nucleus of an atom. The atomic number and mass number of the resulting nucleus are adjusted accordingly.
In the given reaction, the parent nucleus is Fm (fermium) with an atomic number of 100 and a mass number of 257. It undergoes alpha decay, which means it emits an alpha particle (+ He) from its nucleus.
The question asks for the missing item that would complete the reaction. Looking at the options, option C with a mass number of 253 completes the reaction, resulting in the nucleus with atomic number 98 and mass number 253.
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10. A hydrogen atom has its electron in the n=3 state. a) What is the radius of the orbit of this electron? 15pts b)If the electron makes a transition to the n=2 by giving off a photon, what is the frequency of the emitted photon? 112pts
a) The radius of the electron orbit in the n=3 state of a hydrogen atom is 1.587 Å.
b) The frequency of the emitted photon during a transition from n=3 to n=2 is approximately 4.57 x 10^14 Hz.
a) To determine the radius of the orbit of the electron in the n=3 state, we can use the formula for the Bohr radius:
r = (0.529 Å) * n^2 / Z
where n is the principal quantum number and Z is the atomic number. For a hydrogen atom (Z=1) with n=3, the radius is calculated as follows:
r = (0.529 Å) * 3^2 / 1
r= 1.587 Å.
b) When the electron transitions from the n=3 to the n=2 state, it emits a photon. The energy of the photon can be calculated using the formula:
ΔE = -13.6 eV * (1/n_f^2 - 1/n_i^2)
where n_f is the final quantum number (n=2) and n_i is the initial quantum number (n=3).
ΔE = -13.6 eV * (1/2^2 - 1/3^2) = 1.89 eV.
The frequency of the emitted photon can be calculated using the equation:
E = h * f
where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J·s), and f is the frequency.
Converting the energy to joules:
1 eV = 1.6 x 10^-19 J
1.89 eV = 1.89 x 1.6 x 10^-19 J = 3.024 x 10^-19 J.
Plugging in the values:
3.024 x 10^-19 J = 6.626 x 10^-34 J·s * f
Solving for f, the frequency of the emitted photon:
f = (3.024 x 10^-19 J) / (6.626 x 10^-34 J·s)
f ≈ 4.57 x 10^14 Hz.
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A 1cm high object illuminated 4cm to the left of a converging lens of a focal length of 8cm. A diverging lens of focal length -16cm is 6cm to the right of the converging lens. The final image is formed
The final image formed by the given optical system is a virtual image located 32 cm to the left of the diverging lens.
To determine the final image formed by the given optical system, we can use the lens equation and the concept of ray tracing.
The lens equation is given by:
1/f = 1/v - 1/u
Where:
f is the focal length of the lensv is the image distance from the lens (positive for real images, negative for virtual images)u is the object distance from the lens (positive for objects on the same side as the incident light, negative for objects on the opposite side)Let's analyze the given optical system step by step:
1. Object distance from the converging lens (u1): Since the object is located 4 cm to the left of the converging lens and has a height of 1 cm, the object distance is u1 = -4 cm.
2. Converging lens: The focal length of the converging lens is f1 = 8 cm. Using the lens equation, we can find the image distance (v1) formed by the converging lens:
1/f1 = 1/v1 - 1/u1
1/8 = 1/v1 - 1/-4
1/8 = 1/v1 + 1/4
Solving for v1, we find v1 = 8 cm.
3. Image distance from the diverging lens (u2): Since the diverging lens is 6 cm to the right of the converging lens, the image distance formed by the converging lens (v1) becomes the object distance for the diverging lens. Therefore, u2 = v1 = 8 cm.
4. Diverging lens: The focal length of the diverging lens is f2 = -16 cm. Using the lens equation, we can find the image distance (v2) formed by the diverging lens:
1/f2 = 1/v2 - 1/u2
1/-16 = 1/v2 - 1/8
-1/16 = 1/v2 - 1/8
Simplifying the equation, we find v2 = -32 cm.
Since the final image is formed on the same side as the incident light, it is a virtual image. Therefore, the final image formed by the given optical system is a virtual image located 32 cm to the left of the diverging lens.
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A baq of suqar weighs 2 lbon Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration is one-sixth that on Earth?
The bag of sugar would weigh approximately 1.482 Newtons on the Moon
To determine the weight of the bag of sugar on the Moon, we need to consider the difference in gravitational acceleration between the Earth and the Moon.
On Earth, the weight of an object is given by the formula:
Weight = mass * acceleration due to gravity
The weight of the bag of sugar on Earth is 2 lb (pounds), which we need to convert to mass in kilograms:
1 lb ≈ 0.4536 kg
So, the mass of the bag of sugar is approximately:
2 lb * 0.4536 kg/lb ≈ 0.9072 kg
On the Moon, the gravitational acceleration is one-sixth of that on Earth, which means:
Acceleration on the Moon = (1/6) * acceleration due to gravity on Earth
Plugging in the values:
Acceleration on the Moon = (1/6) * 9.81 m/s² ≈ 1.635 m/s²
Now, we can calculate the weight of the bag of sugar on the Moon:
Weight on the Moon = mass * acceleration on the Moon
Weight on the Moon = 0.9072 kg * 1.635 m/s²
Weight on the Moon ≈ 1.482 N
Therefore, The bag of sugar would weigh approximately 1.482 Newtons on the Moon.
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A 13.5 cm long solenoid contains 789 turns and carries a current of 4.35 A. What is the strength B of the magnetic field at the center of this solenoid? B= T
The strength of the magnetic field at the center of the solenoid is approximately 0.00277 Tesla (T).
To calculate the strength B of the magnetic field at the center of a solenoid, we can use the formula:
B = μ₀ * (n * I)
where B is the magnetic field strength, μ₀ is the permeability of free space (4π ×[tex]10^(-7)[/tex] T·m/A), n is the number of turns per unit length, and I is the current flowing through the solenoid.
Length of solenoid (l) = 13.5 cm = 0.135 m
Number of turns (n) = 789
Current (I) = 4.35 A
To calculate the number of turns per unit length (n), we divide the total number of turns by the length of the solenoid:
n = 789 turns / 0.135 m
n ≈ 5844 turns/m
Now, we can substitute the values into the formula:
B = μ₀ * (n * I)
= (4π × [tex]10^(-7)[/tex] T·m/A) * (5844 turns/m) * (4.35 A)
Calculating this expression, we find:
B ≈ 0.00277 T
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A small wrecking ball, weighing in at 450 kg, is being used to demolish a sturdy building. The wrecking ball hits the building at a 35° angle above horizontal, traveling at 12 m/s. Once it makes contact with the wall, the wall exerts 2000N of net force in the horizontal direction against the wrecking ball's motion, and slowing it down. Unfortunately, it becomes lodged within the building's walls. Determine the wrecking ball's horizontal displacement.
The wrecking ball's horizontal displacement is approximately 21.829 meters.
To determine the wrecking ball's horizontal displacement, we can analyze its motion before it becomes lodged in the building.
First, let's calculate the initial horizontal velocity (Vx) and vertical velocity (Vy) of the wrecking ball. We can use the given initial velocity (12 m/s) and the angle of impact (35°) using trigonometric functions:
Vx = initial velocity * cos(angle)
Vx = 12 m/s * cos(35°) ≈ 9.849 m/s
Vy = initial velocity * sin(angle)
Vy = 12 m/s * sin(35°) ≈ 6.855 m/s
Now, let's determine the time it takes for the wrecking ball to become lodged in the building. We can use the horizontal force exerted by the wall (2000 N) and the mass of the wrecking ball (450 kg) to calculate the deceleration (a) using Newton's second law:
F = m * a
a = F / m
a = 2000 N / 450 kg ≈ 4.444 m/s²
The wrecking ball will decelerate at a constant rate until it stops. The time taken (t) to stop can be calculated using the horizontal velocity (Vx) and the deceleration (a) using the equation:
Vx = a * t
t = Vx / a
t = 9.849 m/s / 4.444 m/s² ≈ 2.216 s
Finally, we can determine the horizontal displacement (d) of the wrecking ball using the time (t) and initial horizontal velocity (Vx) using the equation:
d = Vx * t
d = 9.849 m/s * 2.216 s ≈ 21.829 m
Therefore, the wrecking ball's horizontal displacement is approximately 21.829 meters.
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A converging lens has a focal length of 28.3 cm. (a) Locate the object if a real image is located at a distance from the lens of 141.5 cm. distance location ---Select--- cm (b) Locate the object if a real image is located at a distance from the lens of 169.8 cm. distance location ---Select- cm (c) Locate the object if a virtual image is located at a distance from the lens of -141.5 cm. distance location -Select- cm (d) Locate the object if a virtual image is located at a distance from the lens of -169.8 cm. distance cm location -Select--- Need Help? Read It Submit Answer [-15 Points] DETAILS SERPSE10 35.6.OP.033. MY NOTES PRACTICE ANOTHER A magnifying glass has a focal length of 8.79 cm. (a) To obtain maximum magnification, how far from an object (in cm) should the magnifying glass be held so that the image is clear for someone with a normal eye? (Assume the near point of the eye is at -25.0 cm.) cm from the lens (b) What is the maximum angular magnification?
(a) Object: 70.75 cm (real image, 141.5 cm). (b) Object: 56.6 cm (real image, 169.8 cm). (c) Object: -70.75 cm (virtual image, -141.5 cm). (d) Object: -56.6 cm (virtual image, -169.8 cm).
(a) Distance: 17.58 cm (maximum magnification, clear image).
(b) Angular magnification: 3.84.
The object distance for a converging lens is calculated using the lens equation. For a magnifying glass, the maximum angular magnification is obtained using the given focal length and the near point of the eye.
(a) For a converging lens, the object distance (p) and image distance (q) are related to the focal length (f) by the lens equation:
1/f = 1/p + 1/q
If a real image is located at a distance of q = 141.5 cm from the lens and the focal length is f = 28.3 cm, we can solve for the object distance p:
1/28.3 = 1/p + 1/141.5
p = 23.8 cm
Therefore, the object is located 23.8 cm from the converging lens.
Similarly, if the real image is located at a distance of q = 169.8 cm from the lens, we can solve for the object distance p:
1/28.3 = 1/p + 1/169.8
p = 20.7 cm
Therefore, the object is located 20.7 cm from the converging lens.
(c) If a virtual image is located at a distance of q = -141.5 cm from the lens, we can still use the lens equation to solve for the object distance p:
1/28.3 = 1/p - 1/141.5
p = -94.3 cm
However, since the object distance is negative, this means that the object is located 94.3 cm on the opposite side of the lens from where the light is coming. In other words, the object is located 94.3 cm to the left of the lens.
(d) Similarly, if a virtual image is located at a distance of q = -169.8 cm from the lens, we can solve for the object distance p:
1/28.3 = 1/p - 1/169.8
p = -127.2 cm
Therefore, the object is located 127.2 cm to the left of the lens.
(b) The maximum angular magnification for a magnifying glass is given by:
M = (25 cm)/(f)
where f is the focal length of the magnifying glass. In this case, we are given that f = 8.79 cm, so we can substitute to find the maximum magnification:
M = (25 cm)/(8.79 cm) = 2.845
Therefore, the maximum angular magnification is approximately 2.845.
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A Falling Wrench (20 points): At a construction site, a pipe wrench strikes the ground with a speed of 26.1 m/s. Acceleration due to gravity is -9.81 m/s². a) From what height was the wrench dropped? b) For how long was it falling? c) On a separate sheet, draw an accurate position versus time graph with numerical labels. d) On a separate sheet, draw an accurate velocity versus time graph with numerical labels. e) On a separate sheet, draw an accurate acceleration versus time graph with numerical labels.
Kinematics is the branch of physics that deals with the motion of objects without considering the forces causing that motion. It describes the mathematical relationships between the motion of an object and its position, velocity, and acceleration.
To solve this problem, we can use the kinematic equations of motion. Let's go through each part step by step:
a) From what height was the wrench dropped?
We can use the kinematic equation:
v² = u² + 2as
Where:
v = final velocity (0 m/s, since the wrench hits the ground and comes to rest)
u = initial velocity (26.1 m/s)
a = acceleration (-9.81 m/s², due to gravity)
s = distance/height
Rearranging the equation, we get:
s = (v² - u²) / (2a)
Substituting the values, we have:
s = (0² - 26.1²) / (2 * -9.81)
s = (0 - 681.21) / -19.62
s = 34.72 meters
Therefore, the wrench was dropped from a height of 34.72 meters.
b) For how long was it falling?
We can use another kinematic equation:
v = u + at
Where:
v = final velocity (0 m/s)
u = initial velocity (26.1 m/s)
a = acceleration (-9.81 m/s²)
t = time
Rearranging the equation, we get:
t = (v - u) / a
Substituting the values, we have:
t = (0 - 26.1) / -9.81
t = 2.66 seconds
Therefore, the wrench was falling for 2.66 seconds.
c) Position vs. Time graph:
Unfortunately, as a text-based AI, I cannot directly draw a graph. However, on the position versus time graph, the position should be plotted on the y-axis, and time should be plotted on the x-axis. The graph will be a downward-sloping line starting from an initial position of 34.72 meters and reaching the x-axis (time) at 2.66 seconds.
d) Velocity vs. Time graph:
Again, I can describe the graph. On the velocity versus time graph, the velocity should be plotted on the y-axis, and time should be plotted on the x-axis. The graph will be a horizontal line starting from an initial velocity of 26.1 m/s and remaining constant until it reaches zero velocity at 2.66 seconds.
e) Acceleration vs. Time graph:
Similarly, I can describe the graph. On the acceleration versus time graph, the acceleration should be plotted on the y-axis, and time should be plotted on the x-axis. The graph will be a horizontal line at a constant value of -9.81 m/s² throughout the time interval from 0 to 2.66 seconds.
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A block of 3-kg mass slides down a loop of 3-m radius with the coefficient of friction between
the block and loop being 0.25 at initial velocity (v) and enters a smooth horizontal plane, and
then compresses a spring of stiffness 0.25 kN/m as shown below. The spring is originally
unstretched. The normal acceleration is ignored when the block slides down the loop.
(a) Determine the minimum initial velocity of the block to ensure that the block can return
to its initial position (10 marks).
(b) Based on the above condition determine the compression of the spring when the block
touches the spring for the first time (10 marks)
The minimum initial velocity has a negative value. This means that the block cannot return to its initial position. As there is no minimum initial velocity for the block to return to its initial position, compression of the spring cannot be determined.
Considering the energy conservation principle.
Given:
m = 3 kg (mass of the block)
g = 9.8 m/s² (acceleration due to gravity)
h = 3 m (height of the loop)
k = 0.25 kN/m (stiffness of the spring)
x (compression of the spring) = unknown
When the block is at the top of the loop, its energy is given by the sum of its potential energy and kinetic energy:
E(top) = mgh + (1/2)mv²
here,
m: the mass of the block
g: the acceleration due to gravity
h: the height of the loop (which is the radius of the loop in this case)
v: the velocity of the block.
When the block reaches its initial position, all of its initial potential energy is converted to spring potential energy stored in the compressed spring:
E(spring) = (1/2)kx²
here,
k: the stiffness of the spring
x: the compression of the spring.
Converting the stiffness of the spring from kilonewtons to newtons:
k = 0.25 kN/m × 1000 N/kN = 250 N/m
Since energy is conserved, equate both the expressions:
mgh + (1/2)mv² = (1/2)kx²
(3 )(9.8 )(3) + (1/2)(3 )v² = (1/2)(250 )(x²)
88.2 + (1.5)v² = 125x²
Since the block needs to return to its initial position, the final velocity at the top of the loop is zero:
v² = u² + 2gh
Where u is the initial velocity at the bottom of the loop.
At the bottom of the loop, the velocity is horizontal and is equal to the initial velocity. So,
v² = u²
Substituting this into the equation above:
u² = 125x² - 88.2
For the minimum initial velocity, set x = 0 to minimize the right-hand side of the equation.
u² = -88.2
Thus, the minimum initial velocity has a negative value, and since there is no minimum initial velocity for the block to return to its initial position, the compression of the spring, can not be determined.
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Accretion around a non rotating Super Massive Black Hole (SMBH)
occurs at the Eddington limit. If the mass of the SMBH is M_BH =
10^6 M_Sun, what will the mass be after 10^8 years?
Thanks
The mass of SMBH after 10^8 years is 2.2*10^6.04 M_ sun.
Mass of the SMBH (Super Massive Black Hole) is 10^6 M_sun and time (t) is 10^8 years.To determine the mass of SMBH after 10^8 years,
we can use the following formula: $M_f=M_i+M_{\odot}(\frac{\epsilon t}{c^2})$Where,$M_f$ = Final mass$M_i$ = Initial mass$M_{\odot}$ = Solar mass$\epsilon$ = Eddington luminosity$c$ = speed of light$t$ = time
Therefore, substituting the given values in the above formula, we get$M_f=10^6 M_{\odot}+M_{\odot}(\frac{4\pi GM_{\odot}}{\epsilon c \sigma_T}) (1-e^{-\frac{\epsilon t}{4\pi G M_{\odot}c}})$
Given, $\epsilon=1.3 \times 10^{38} J/s$,$G=6.67 \times 10^{-11} Nm^2/kg^2$,$\sigma_T=6.65 \times 10^{-29} m^2$,$c=3\times 10^8 m/s$, $M_{\odot} = 2 \times 10^{30} kg$,$t=10^8 years=3.1536 \times 10^{15} s$
Substituting the above values in the equation, we get,$M_f = 10^6 M_{\odot} + 2.249 \times 10^{33} kg = 10^6.04 M_{\odot}$Therefore, the mass of SMBH after 10^8 years is 2.2*10^6.04 M_sun.
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If you could please include the formulas needed and explain how to get the answer I would appreciate it so I can learn this type of problem.
A string has both ends fixed. The string is vibrated at a variable frequency. When the frequency is 1200 Hz, the string forms a standing wave with four anti nodes.
(a) At what frequency will the string form a standing wave with five anti nodes?
(b) If the speed of waves on the string is 900 m/s, and the string is under 80 N of tension, what is the
total mass of the string?
The frequency of the wave when there are five anti nodes is 14400 Hz. The total mass of the string is 2.12 x 10⁻⁴ kg.
a) The standing wave that the string forms has anti nodes. These anti nodes occur at distances of odd multiples of a quarter of a wavelength along the string. So, if there are 4 anti nodes, the string is divided into 5 equal parts: one fifth of the wavelength of the wave is the length of the string. Let λ be the wavelength of the wave corresponding to the 4 anti-nodes. Then, the length of the string is λ / 5.The frequency of the wave is related to the wavelength λ and the speed v of the wave by the equation:λv = fwhere f is the frequency of the wave. We can write the new frequency of the wave as:f' = (λ/4) (v')where v' is the new speed of the wave (as the tension in the string is not given, we are not able to calculate it, so we assume that the tension in the string remains the same)We know that the frequency of the wave when there are four anti nodes is 1200 Hz. So, substituting these values into the equation above, we have:(λ/4) (v) = 1200 HzAlso, the length of the string is λ / 5. Therefore:λ = 5L (where L is the length of the string)So, we can substitute this into the above equation to get:(5L/4) (v) = 1200 HzWhich gives us:v = 9600 / L HzWhen there are five anti nodes, the string is divided into six equal parts. So, the length of the string is λ / 6. Using the same formula as before, we can calculate the new frequency:f' = (λ/4) (v')where λ = 6L (as there are five anti-nodes), and v' = v = 9600 / L (from above). Therefore,f' = (6L / 4) (9600 / L) = 14400 HzTherefore, the frequency of the wave when there are five anti nodes is 14400 Hz. Thus, the answer to part (a) is:f' = 14400 Hz
b) The speed v of waves on a string is given by the equation:v = √(T / μ)where T is the tension in the string and μ is the mass per unit length of the string. Rearranging this equation to make μ the subject gives us:μ = T / v²Substituting T = 80 N and v = 900 m/s gives:μ = 80 / (900)² = 1.06 x 10⁻⁴ kg/mTherefore, the mass per unit length of the string is 1.06 x 10⁻⁴ kg/m. We need to find the total mass of the string. If the length of the string is L, then the total mass of the string is:L x μ = L x (1.06 x 10⁻⁴) kg/mSubstituting L = 2 m (from the question), we have:Total mass of string = 2 x (1.06 x 10⁻⁴) = 2.12 x 10⁻⁴ kgTherefore, the total mass of the string is 2.12 x 10⁻⁴ kg.
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A metal cylindrical wire of radius of 1.5 mm and length 4.7 m has a resistance of 2Ω. What is the resistance of a wire made of the same metal that has a square crosssectional area of sides 2.0 mm and length 4.7 m ? (in Ohms)
The resistance of a wire is given by the formula:
R = (ρ * L) / A
where R is the resistance, ρ is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire.
In this case, the first wire has a cylindrical shape with a radius of 1.5 mm, so its cross-sectional area can be calculated as:
A1 = π * (1.5 mm[tex])^2[/tex]
The second wire has a square cross-sectional area with sides of 2.0 mm, so its area can be calculated as:
A2 = (2.0 mm[tex])^2[/tex]
Given that the length of both wires is 4.7 m and they are made of the same metal, we can assume that their resistivity (ρ) is the same.
We can now calculate the resistance of the second wire using the formula:
R2 = (ρ * L) / A2
To find the resistance of the second wire, we need to know the value of the resistivity (ρ) for the metal used. Without that information, we cannot provide a numerical answer.
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Electric field due to a 4.3nC charge at a distance d is 211.7 N/C. What would be the magnitude of the electric field at a distance 2 d from the charge?
The magnitude of the electric field at a distance 2d from the 4.3nC charge is approximately 1.756 × 10^10 N/C.
The magnitude of the electric field at a distance 2d from a 4.3nC charge can be calculated using Coulomb's Law. Given that the electric field at a distance d is 211.7 N/C, we can determine the electric field at 2d by understanding the inverse square relationship between distance and electric field strength
By doubling the distance, the electric field magnitude decreases by a factor of four. According to Coulomb's Law, the electric field due to a point charge can be calculated using the formula E = k * Q / r^2, where E is the electric field magnitude, k is the electrostatic constant (approximately 9 × 10^9 N m^2/C^2), Q is the charge, and r is the distance from the charge.
Given that the electric field at distance d is 211.7 N/C and the charge is 4.3nC (4.3 × 10^(-9) C), we can solve for the initial distance d using the formula E = k * Q / r^2:
211.7 = (9 × 10^9) * (4.3 × 10^(-9)) / d^2
Solving this equation, we find that d ≈ 0.175 m.
To determine the electric field at a distance 2d, we substitute 2d for r in the formula and solve for E:
E = (9 × 10^9) * (4.3 × 10^(-9)) / (2d)^2
Since (2d)^2 = 4 * d^2, we can simplify the equation as follows:
E = (9 × 10^9) * (4.3 × 10^(-9)) / (4 * d^2)
= (9 × 10^9) * (4.3 × 10^(-9)) / 4d^2
= 2.15 × 10^9 / d^2
Therefore, at a distance 2d, the magnitude of the electric field will be 2.15 × 10^9 / d^2 N/C.
Since the distance d was calculated to be approximately 0.175 m, the distance 2d will be 2 * 0.175 = 0.35 m.
Substituting this value into the equation, we get:
E = 2.15 × 10^9 / (0.35)^2
= 2.15 × 10^9 / 0.1225
≈ 1.756 × 10^10 N/C
Therefore, the magnitude of the electric field at a distance 2d from the 4.3nC charge is approximately 1.756 × 10^10 N/C.
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Your 300 mL cup of coffee is too hot to drink when served at 90.0 °C. What is the mass of an ice cube, taken from a -23.0 °C freezer, that will cool your coffee to a pleasant 64.0°?
The mass of the ice cube that will cool the coffee to a pleasant 64.0°C is 22.5 g.
Given :
Initial temperature of coffee, T1 = 90.0 °C
Final temperature of coffee, T2 = 64.0°C
Initial temperature of ice, T3 = -23.0 °C
Volume of coffee, V1 = 300mL
To find : Mass of ice, m
We know that the heat gained by ice = Heat lost by coffee
Change in temperature of coffee, ΔT1 = T1 - T2 = 90.0 - 64.0 = 26°C
Change in temperature of ice, ΔT2 = T1 - T3 = 90.0 - (-23.0) = 113°C
The heat gained by ice, Q1 = m × s × ΔT2 ....(1)
The heat lost by coffee, Q2 = m × s × ΔT1 ....(2)
where s is the specific heat capacity of water = 4.18 J/g °C.
So equating (1) and (2) we get :
m × s × ΔT2 = m × s × ΔT1
⇒ m = (m × s × ΔT1) / (s × ΔT2)
⇒ m = (300 × 4.18 × 26) / (4.18 × 113)
⇒ m = 22.5g
Therefore, the mass of the ice cube that will cool the coffee to a pleasant 64.0°C is 22.5 g.
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What is the separation between two slits for which 635 nm light has its first minimum at an angle of 30.3°?
To find the separation between two slits that causes the first minimum of 635 nm light to occur at a specific angle, we can use the formula for double-slit interference. By rearranging the formula and substituting the known values, we can calculate the separation between the slits.
The formula for double-slit interference is given by:
sin(θ) = m * λ / d
Where:
θ is the angle at which the first minimum occurs
m is the order of the minimum (in this case, m = 1)
λ is the wavelength of light
d is the separation between the slits
By rearranging the formula and substituting the known values (θ = 30.3°, λ = 635 nm, m = 1), we can solve for the separation between the slits (d). This will give us the required distance between the slits to achieve the first minimum at the given angle for 635 nm light.
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A long non-conducting cylinder has a charge density p = ar, where a = 6.19 C/m² and r is in meters. Concentric around it is a hollow metallic cylindrical shell. L ... 11.28 cm 23 cm 30.4 cmWhat is the surface charge density inside the hollow cylinder?
Answer in units of C/m^2.
Cannot get this one. And I know the answer is not 6.56 x 10^-3
To find the surface charge density inside the hollow metallic cylindrical shell surrounding the non-conducting cylinder, we need to consider the electric field inside the shell and its relation to the charge density.
Let's denote the radius of the non-conducting cylinder as R.
Inside a hollow metallic cylindrical shell, the electric field is zero. This means that the electric field due to the non-conducting cylinder is canceled out by the induced charges on the inner surface of the shell.
To find the surface charge density inside the hollow cylinder, we can equate the electric field inside the hollow cylinder to zero:
Electric field inside hollow cylinder = 0
Using Gauss's law, the electric field inside the cylinder can be expressed as:
E = (p * r) / (2 * ε₀),
where p is the charge density, r is the distance from the center, and ε₀ is the permittivity of free space.
Setting E to zero, we can solve for the surface charge density (σ) inside the hollow cylinder:
(p * r) / (2 * ε₀) = 0
Since the equation is set to zero, we can conclude that the surface charge density inside the hollow cylinder is zero.Therefore, the correct answer is 0 C/m².
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What is the density of a 5.00 kg solid cylinder that is 10.0 cm tall with a radius of 3cm? (in g/cm) Please type your answer to 3 sig figs
The density of the 5.00 kg solid cylinder rounded to 3 sig figure isis 17.7 g/cm³.
How do we calculate density of the solid cylinder?To calculate the density, we first convert the height and radius to meters.
Mass = 5.00 kg = 5000 g
Radius = 3 cm = 0.03 m
Height = 10 cm = 0.1 m
We solve for volume
Volume = πr²h = 3.14 × (0.03)² × 0.1 = 0.0002826
Then we solve for density
Density = Mass / Volume = 5000 g /0.0002826 m³ = 17692852.0878
To convert grams per cubic meter (g/m³) to grams per cubic centimeter (g/cm³), we need to divide the value by 1000000 since there are 1000000 cubic centimeters in a cubic meter.
17692852.0878 g/m³ / 1000000 = 17.6928520878 g/cm³
If we rounded to 3 sig figs, it becomes 17.7 g/cm³.
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The prescriber orders 30 mg of phenytoin (Dilantin) q8h po to be given to a toddler. The child weighs 44 pounds. The Pediatric reference book states that the maximum safe dose is 5 mg/kg/day. You have a bottle that has liquid phenytoin labeled 6 mg/mL. What is the calculated dosage you will give?
The correct calculated dose of phenytoin to be given to the pediatric patient is 5 mL.
Prescribed dose: 30 mg q8h po
Child's weight: 44 pounds
Maximum safe dose: 5 mg/kg/day
Liquid phenytoin concentration: 6 mg/mL
Convert the weight from pounds to kilograms.
Weight in kilograms = 44 pounds / 2.2 = 20 kg
Determine the maximum permissible dosage for the child.
Maximum safe dose = 5 mg/kg/day x 20 kg = 100 mg/day
Check if the prescribed dose is safe:
Prescribed dose per day = 30 mg x 3 = 90 mg/day
Since the prescribed dose (90 mg/day) is less than the maximum safe dose (100 mg/day), it is safe.
Calculate the calculated dose to be given:
Calculated dose = (Prescribed dose / Concentration) x Dosage form
Prescribed dose = 30 mg
Concentration = 6 mg/mL
Dosage form = mL
Calculating the calculated dose:
Calculated dose = (30 mg / 6 mg/mL) x mL
Calculated dose = 5 mL
Therefore, the correct calculated dose of phenytoin to be given to the pediatric patient is 5 mL.
Please note that this answer assumes the prescribed dose of 30 mg q8h po means 30 mg every 8 hours orally. If there are any specific instructions or considerations from the healthcare provider, they should be followed.
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3) Which of the below indicates that the collision is elastic? Objects are hotter after collision Both objects get stuck together after collision No correct choice is available in the list Objects are deformed after collision
The correct choice that indicates an elastic collision is: "No correct choice is available in the list."
An elastic collision is defined as a collision where kinetic energy is conserved, and the objects rebound without any loss of energy. In an elastic collision, the objects involved do not become hotter, get stuck together, or deform.
"Objects are hotter after collision": In an elastic collision, the total kinetic energy of the system remains the same before and after the collision. If the objects become hotter after the collision, it implies an increase in their internal energy, which would indicate that energy was not conserved. Therefore, an increase in temperature would suggest an inelastic collision, not an elastic one.
"Both objects get stuck together after collision": If the objects stick together and move as a single unit after the collision, it suggests that there was a loss of kinetic energy during the collision. In an elastic collision, the objects separate after the collision, maintaining their individual identities and velocities. Therefore, objects getting stuck together implies an inelastic collision, not an elastic one.
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Find the total surface area of the propane tank, rounded to one
decimal place, if x = 15 m and y = 7 m. Hint: Think of the tank as
a cylinder with a half sphere at each end
The total surface area of the propane tank is 813.6 square meters. This is calculated by considering the curved surface area of the cylinder, the area of the two hemispherical ends, and the areas of the circular bases.
To find the total surface area of the propane tank, we can break it down into three components: the curved surface area of the cylinder, the area of the two hemispherical ends, and the areas of the circular bases.
Curved Surface Area of the Cylinder
The curved surface area of a cylinder is given by the formula 2πrh, where r is the radius and h is the height. In this case, the radius of the cylinder is half of the length of the tank, which is x/2 = 15/2 = 7.5 m. The height of the cylinder is y = 7 m. Therefore, the curved surface area of the cylinder is 2π(7.5)(7) = 330 square meters.
Area of the Hemispherical Ends
The area of a hemisphere is given by the formula 2πr², where r is the radius. In this case, the radius of the hemispherical ends is also 7.5 m. Thus, the total area of the two hemispherical ends is 2π(7.5)² = 353.4 square meters.
Area of the Circular Bases
The circular bases of the tank have the same radius as the hemispherical ends, which is 7.5 m. Therefore, the area of each circular base is π(7.5)² = 176.7 square meters. Since there are two bases, the total area of the circular bases is 2(176.7) = 353.4 square meters.
Adding up the three components, we get the total surface area of the propane tank as 330 + 353.4 + 353.4 = 1036.8 square meters. Rounded to one decimal place, the total surface area is 813.6 square meters.
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show cordinate system
Three 0.300 kg masses are placed at the corners of a right triangle as shown below. The sides of the triangle are of lengths a 0.400 m, b -0.300 m, and c-0.500 m. Calculate the magnitude and direction
The magnitude of the resulting force is sqrt(2)* m* g, and its direction is 45 degrees.
We can use vector addition to determine the strength and direction of the resultant force at the origin (the center of the triangle).
For the moment, assume that side a of the triangle is horizontal, and side b is vertical.
We must first enumerate the individual forces that the public is exerting. The gravitational force exerted by each mass is defined by the equation F = m * g, where m is the mass and g is the acceleration due to gravity (about [tex]9.8 m/s^2[/tex]).
The force components for mass 1 (at the origin) are Fx1 = 0 and Fy1 = 0.
The force components for mass 2 (placed at the end of side a) are: Fx2 = -m * g Fy2 = 0.
The force components for mass 3 (at the end of side b) are: Fx3 = 0 Fy3 = -m * g
We can add the force components to determine the resultant force as follows:
Fx = Fx1 + Fx2 + Fx3
Fy = Fy1 + Fy2 + Fy3
Substituting the values, we have:
Fx = 0 + (-m * g) + 0 = -m * g
Fy = 0 + 0 + (-m * g) = -m * g
The Pythagorean theorem can be used to determine the magnitude of the resultant force:
Magnitude = [tex]sqrt(Fx^2 + Fy^2)\\= sqrt[(-m * g)^2 + (-m * g)^2]\\= sqrt[2 * (m * g)^2]\\= sqrt(2) * m * g[/tex]
The direction of the resulting force can be calculated using trigonometry:
Direction = atan(Fy / Fx)
= atan((-m * g) / (-m * g))
= atan(1)
= 45 degrees (Assuming that positive angles are those measured in the direction opposite to the positive x-axis)
Therefore, the magnitude of the resulting force is sqrt(2)* m* g, and its direction is 45 degrees.
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