the pressure of the water after the contraction is -50000 Pa (or 50 kPa below atmospheric pressure), and the velocity of the water after the contraction is 15.0 m/s.
The continuity equation states that the product of the cross-sectional area and the velocity of an incompressible fluid is constant along a pipe, so we can use it to relate the pressure and velocity before and after the contraction:
A₁v₁ = A₂v₂
where A₁ and v₁ are the area and velocity of the pipe before the contraction, and A₂ and v₂ are the area and velocity of the pipe after the contraction.
We can also use the Bernoulli equation, which relates the pressure and velocity of a fluid along a streamline:
P₁ + 1/2 ρv₁² = P₂ + 1/2 ρv₂²
where P₁ and v₁ are the pressure and velocity of the fluid before the contraction, and P₂ and v₂ are the pressure and velocity of the fluid after the contraction, and ρ is the density of the fluid, which we assume to be constant.
Solving for the pressure and velocity after the contraction, we can use the continuity equation to express v₁ in terms of v₂ and substitute it into the Bernoulli equation:
A₁v₁ = A₂v₂
v₁ = (A₂/A₁) v₂
P₁ + 1/2 ρ((A₂/A₁) v₂)² = P₂ + 1/2 ρv₂²
Simplifying and solving for P₂, we get:
P₂ = P₁ + 1/2 ρ(v₁² - v₂²)
Substituting the given values, we get:
A₂ = (1/3) A₁
v₁ = 5.0 m/s
P₁ = 450000 Pa
ρ = 1000 kg/m³
Using the continuity equation, we can find the value of v₂:
A₁v₁ = A₂v₂
v₂ = (A₁/A₂) v₁
v₂ = 3 × 5.0 m/s
v₂ = 15.0 m/s
Substituting this value into the Bernoulli equation, we can find the pressure P₂:
P₂ = P₁ + 1/2 ρ(v₁² - v₂²)
P₂ = 450000 Pa + 1/2 × 1000 kg/m³ × (5.0 m/s)² - (15.0 m/s)²
P₂ = 450000 Pa - 500000 Pa
P₂ = -50000 Pa
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An object of mass 0.35kg oscillates in SHM with an amplitude of 140mm
and a frequency of 0.60Hz.
Calculate or find:
i) Maximum kinetic energy of the object.
ii) Maximum potential energy of the object.
iii) Potential and kinetic energy at the mid-way point between the centre and the
extremity of the motion.
Answer:
The equation for the total mechanical energy of an object in SHM is:
E = 1/2 kA^2
where E is the total mechanical energy, k is the spring constant, and A is the amplitude.
To solve the problem, we need to find the spring constant of the oscillator:
f = 1/T
where f is the frequency and T is the period.
T = 1/f = 1/0.60 = 1.67 s
The angular frequency of the oscillator is:
ω = 2πf = 2π/T = 3.76 rad/s
The spring constant of the oscillator is:
k = mω^2 = 0.35 x (3.76)^2 = 4.97 N/m
i) The maximum kinetic energy of the object is equal to the maximum potential energy, which is:
Emax = 1/2 kA^2 = 1/2 x 4.97 x (0.14)^2 = 0.012 J
ii) The maximum potential energy of the object is the same as the maximum kinetic energy, which is:
Emax = 1/2 kA^2 = 1/2 x 4.97 x (0.14)^2 = 0.012 J
iii) At the mid-way point between the centre and the extremity of the motion, the displacement of the oscillator is half the amplitude, which is 70 mm or 0.07 m. At this point, the kinetic energy is zero, and the potential energy is:
E = 1/2 kx^2 = 1/2 x 4.97 x (0.07)^2 = 0.012 J
Therefore, the total mechanical energy at this point is also 0.012 J.
The Earth surface temperature is around 270 K and emissivity of 0.8, while space has temperature of around 2K. What is the net power radiated by the Earth in free space?
The net power radiated by the Earth in free space is approximately
1.2 x 10^ 17 W watts.
How to find the power radiatedTo calculate the net power radiated by the Earth in free space, we can use the Stefan-Boltzmann Law:
Power = σ * A * ε * (T^4 - T0^4)
where:
σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4)
A is the surface area of the Earth (4πR^2, where R is the radius of the Earth)
ε is the emissivity of the Earth (0.8)
T is the temperature of the Earth surface (270 K)
T0 is the temperature of space (2 K)
Plugging in the values, we get:
Power = 5.67 x 10^-8 * 4πR^2 * 0.8 * (270^4 - 2^4)
The radius of the Earth is approximately 6.37 x 10^6 m, so:
Power = 5.67 x 10^-8 * 4π(6.37 x 10^6)^2 * 0.8 * (270^4 - 2^4)
Power ≈ 1.193 x 10^ 17 W
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Need help I don't know if these are label correct but I need to know which letter on the diagram correctly labels amplitude of the wave and please label the rest
Pic attached below
Explanation:
I wavelength
J amplitude
K trough
L crest
M equal
Which statements are true about the amount of gravitational potential energy an object has?
The amount increases as the object is lifted higher.
The amount varies according to the material of the object.
The amount varies according to the mass of the object.
The amount increases the more quickly the object is lifted.
Answer:
The statement "The amount increases as the object is lifted higher" is true about the amount of gravitational potential energy an object has. The statement "The amount varies according to the material of the object" is false. The amount of gravitational potential energy only depends on the mass of the object and its elevation from the reference point. The statement "The amount varies according to the mass of the object" is true. The more massive an object is, the more gravitational potential energy it has. The statement "The amount increases the more quickly the object is lifted" is false. The amount of gravitational potential energy only depends on the object's height above the reference point, not the speed at which it is lifted.
Answer: (A)
Explanation: The amount increases as the object is lifted higher.
for the circuit shown in find the output voltage(ii) the current through zenor diode
A Zener diode is a heavily made semiconductor device that is designed to operate in the reverse direction.Voltage drop across series resistance is 70 volts and the current through zenor diode is 9mA.
When the voltage across the terminals of a diode is reversed, and the potential reaches the voltage (knee voltage), the junction break down, and the current flows in the reverse direction. This effect is known as the diode effect.
R = 5K ohms = 5 x 10^3 ohms
Input voltage = 12V, Zener voltage = 50V
Output Voltage = 50V
Voltage drop across series resistance = input voltage – zener voltage = 120 – 50 = 70 volts
Load Current = zener voltage / resistance = \frac{50}{10 x 10^3} = 5 x 10^{-3} A
Current through = \frac{input voltage – zener voltage} {resistance}
= 70 / 5 x 10^{-3} = 14 x 10^{-3} A
According to kirchoff’s first law = current + zener current
Zener current zener current = I - line current = 14 x 10^{-3} - 5 x 10^{-3} = 9 x 10^{-3} = 9 mA
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How strong would a magnetic field need to be in order to make a particle with a mass of 4 x 10−11 kg and a charge of 8 nC move in a circular path with a speed of 400 m/s and a radius of 0.5 m?
As a result, a particle with a mass of 4 x 10-11 kg and a charge of 8 nC needs a magnetic field of 1.0 Tesla to move in a circular path with a speed of 400 m/s and a radius of 0.5 m.
When is the strongest magnetic pressure acting on a charged particle in a magnetic area?Thus, when a charged particle moves at a 90° angle to the field, the magnetic force acting on it is greatest.
In a magnetic subject, what is the pressure on a moving charge?What force does a moving object experience in a magnetic field? As a charge moves through an electric and magnetic field, the Lorentz force acts on it. The total of magnetic and electric forces must be what it is.
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A metal ball bearing with mass 5.0 g falls out of a factory machine and drops to the concrete floor 3.0 m below. It bounces back up to its starting point. Find the changes in the bearing's potential and kinetic energies as it a) travels from the machine down to the floor, and b) travels up from the floor back to its starting point.
a) When the metal ball bearing falls from the machine to the concrete floor, its potential energy decreases while its kinetic energy increases.
The potential energy lost by the ball bearing can be calculated using the formula PE = mgh, where m is the mass of the ball bearing (0.005 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the height the ball bearing falls (3.0 m). Thus, the potential energy lost by the ball bearing is:
PE = (0.005 kg)(9.8 m/s²)(3.0 m) = 0.147 J
At the same time, the ball bearing's kinetic energy increases by an amount equal to the potential energy lost. Therefore, the ball bearing's initial kinetic energy is zero, and its final kinetic energy is:
KE = 0.147 J
b) As the ball bearing bounces back up from the concrete floor to its starting point, its kinetic energy decreases while its potential energy increases. The kinetic energy lost by the ball bearing can be calculated as the same value as before:
KE = 0.147 J
The ball bearing's potential energy at its starting point is equal to the potential energy it lost on the way down, which is:
PE = 0.147 J
Therefore, the ball bearing's change in potential energy is 0.147 J (from down to up), and its change in kinetic energy is -0.147 J (from up to down).
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Answer: a) When the ball bearing falls from the machine to the floor, there is a change in its potential and kinetic energies. The potential energy of an object at a height h above the ground is given by mgh, where m is the mass of the object, g is the acceleration due to gravity (9.8 m/s^2), and h is the height above the ground. Initially, the ball bearing is at rest on the machine, so its kinetic energy is zero. Therefore, the initial energy of the ball bearing is purely potential energy, given by:
PEi = mgh = (0.005 kg)(9.8 m/s^2)(3.0 m) = 0.147 J
When the ball bearing hits the ground, its potential energy is zero and its kinetic energy is at a maximum. The velocity of the ball bearing just before it hits the ground can be found using the equation:
v^2 = u^2 + 2as
where u is the initial velocity (which is zero), a is the acceleration due to gravity (-9.8 m/s^2), s is the distance fallen (3.0 m), and v is the final velocity just before hitting the ground. Solving for v, we get:
v = sqrt(2as) = sqrt(2*(-9.8 m/s^2)*(3.0 m)) = 7.67 m/s
The kinetic energy of the ball bearing just before it hits the ground is given by:
KEf = (1/2)mv^2 = (1/2)(0.005 kg)(7.67 m/s)^2 = 0.145 J
Therefore, the change in potential energy is:
ΔPE = PEf - PEi = 0 - 0.147 J = -0.147 J
And the change in kinetic energy is:
ΔKE = KEf - KEi = 0.145 J - 0 J = 0.145 J
b) When the ball bearing bounces back up to its starting point, there is another change in its potential and kinetic energies. Just before it reaches its highest point, the ball bearing's velocity is zero, so its kinetic energy is also zero. Therefore, its energy is purely potential energy, given by:
PEf = mgh = (0.005 kg)(9.8 m/s^2)(3.0 m) = 0.147 J
The ball bearing reaches its highest point when all of its initial kinetic energy has been converted to potential energy. At this point, its potential energy is at a maximum and its kinetic energy is at a minimum. The ball bearing then starts to fall back down towards the ground, and its potential energy starts to decrease while its kinetic energy increases. Just before it hits the ground, its kinetic energy is at a maximum and its potential energy is at a minimum, as we saw in part (a). When it bounces back up, the process repeats.
Therefore, the change in potential energy as the ball bearing travels from the floor back up to its starting point is:
ΔPE = PEf - PEi = 0.147 J - 0 J = 0.147 J
And the change in kinetic energy is:
ΔKE = KEf - KEi = 0 J - 0 J = 0 J
Note that the change in kinetic energy is zero because the ball bearing starts and ends at rest.
Explanation: can i get brainliest
List 3 components of an electric circuit.
Answer:
An energy source – like a battery or mains power.
An energy receiver – like a lightbulb.
An energy pathway – like a wire.
Explanation:
A uniform line of charge with length 20.0 cm is along the x-axis, with its midpoint at x =0. Its charge per length is +5.30 nC/m. A small sphere with charge -4.00 μC is located at x= 0, y = 5.00 cm.
A) What is the magnitude of the force that the charged sphere exerts on the line of charge?
B) What is the direction angle of the force that the charged sphere exerts on the line of charge? The angle is measured from the +x-axis toward the +y-axis.
The charged spheroid pushes against the line of charge with a force of 0.0181 N.
How much of a power is acting on the charge?Its magnitude is determined by the Coulomb's law, F = k q1 q2/r2, when a point charge (a particle with a charge Q) is operating on a test charge q at a distance r.
Force between the ions' magnitude
Apply the electrostatic force equation of Coulomb.
F = (kq1q2)/r²
where;
q1 is charge 1=6.3nC/mx0.2m=1.26 nC
q2 is charge 2=-4 μC
r is the distance between the charges=5 cm= 0.05 m
F = (9x10⁹x1.26x10⁻⁹x4x10⁻⁶)/(0.05)²
F = 0.0181 N.
B) By breaking the force down into its x and y components, you can determine the orientation of the force the charged sphere applies to the line of charge.
θ = arctan(Fy/Fx) = arctan(Fy/0) = arctan(-Fy)
where Fy is the force's y-component. The force's y component can be calculated as follows:
Fy = F * sin(θ) = F * sin(arctan(-Fy))
Solving for Fy, we get:
Fy = -F * sin(arctan(-Fy)) = -F * (-0.05 / r) = 0.05F / √(x² + 0.05²)
where we used the fact that sin(arctan(x)) = x/√(1+x²).
Plugging in the values, we get:
Fy = 0.05 * 4.47 × 10⁻⁴/ √(0² + 0.05²)
≈ 4.00 × 10⁻⁵ N
Therefore, the direction angle is:
θ = arctan(-Fy/Fx) = arctan(4.00 × 10⁻⁵/ 0) = 90°
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Two +1 C charges are separated by 300m. What is the magnitude of the electric force between them
Answer:
1000 N
Explanation:
The electric force between two charges is given as
F = kq'q/r²......................... Equation 1
Where F = electric force between the charges, q' = magnitude of the first charge, q = magnitude of the second charge, r = distance between the charges, k = coulombs constant
Given: q' = q = 1 C, r = 3000 m, k = 9×10⁹ Nm²/kg²
Substitute into equation 2
F = 1²×(9×10⁹)/3000²
F = 1000 N.
Hence the magnitude of the electric force between them = 1000 N
2. What gravitational force does the moon produce on the Earth if their centers are 3.88 x10^8 m apart and the moon has a mass of 7.43 x 10^22 kg?
About 1.98 x 10²⁰ Newtons of gravitational force are exerted by the moon on the Earth.
The moon is located 3.84 x 10⁸ metres away from the earth.The angle subtended if observed from two diametrically opposed places on the Earth. The moon is 3.84 10⁸ metres away from the Earth. The angle subtended at the moon, when seen from two diametrically opposed places on Earth, is 1° 54′.
where F is the gravitational force, G is the gravitational constant (6.6743 x 10⁻¹¹ N*m²/kg²), m1 and m2 are the masses of the two objects, and r is the distance between their centers.
In this case, we are calculating the force that the moon produces on the Earth, so we can set m1 to the mass of the Earth (5.97 x 10²⁴ kg).
F = 6.6743 x 10⁻¹¹ * (5.97 x 10²⁴ kg) * (7.43 x 10²² kg) / (3.88 x 10⁸ m)²
Simplifying the calculation, we get:
F = 1.98 x 10²⁰ N
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easy physics. HELP
if a cart is carrying 100kg of mass, at what rate will the cart accelerate if 200 N of force are applied? (a=F/m)
a. 20m/s *2
b. 50 m/s *2
c. 2m/s *2
d. 100 m/s *2
8. A man with an open parachute falls to Earth at constant speed. The following forces act: 9 P the upward force of the parachute on the man Q the upward force of the man on the Earth R the downward force of the Earth on the parachute S the downward force of the man on the parachute. Which two forces are a Newton's third law pair? A. P and Q 9 ● B. P and R C. P and S D. Q and R
Answer:
Explanation:
According to Newton's third law, for every action, there is an equal and opposite reaction. This means that if force A acts on object B, then there must be a force that object B exerts on object A that is equal in magnitude but opposite in direction. Therefore, the two forces that are a Newton's third law pair are P and Q, as they are the forces that the parachute and the man exert on each other in opposite directions.
R and S are not a Newton's third law pair because they are not acting on the same objects. R is the force of the Earth on the parachute, while S is the force of the man on the parachute. These forces are not equal and opposite, as they are acting on different objects.
A 1.3 KG blocks flies along a frictionless surface at 1.0 M/S.a2 block sliding at a faster 5.0 M/S collides with the first from behind and sticks to it. The final velocity of the combine blocks is 2.0 M/S. What was the mass of the second block?
the initial momentum of the system of block m1 and block m2 is
Pi= m1v1 + m2v2
the final momentum of the combine blocks is
Pf= (m1+m2)V
according to the law of convervation of momentum
Pi = Pf
m1v1 + m2v2 = (m1+m2)V
1.3 × 1 + 5m2 = 1.3 × 2 + 2m2
m2= 1.3/3 kg
A rectangle with a base labeled uppercase L and a height labeled lowercase w. A vertical, dashed line runs along the leftside of the rectangle's height. A circular arrow around the vertical, dashed line indictates that the rectangle rotates about its left edge.
Calculate the moment of inertia if the plate has a length of 9.00 cm, a width
of 7.00 cm, and a uniform mass density of 2.50 g/cm^2
A = wL = 63 cm²
m = (2.5 g/cm²)(63 cm²)
m = 157.5 g = 0.1575kg
[tex]I=\frac{1}{3} mL^2\\\\I = \frac{1}{3}(0.1575kg)(0.09m)^2[/tex]
I = 4.2525×10⁻⁴ kg/m²
the velocity of three turtles walking in the park. The x-axis is positive towards the North direction.
1. Rank the accelerations of the three turtles from the smallest to largest
2. Are the turtles go- ing towards North or South direction? Are they changing direction?
The ranking of their accelerations from smallest to largest is: Turtle C, Turtle A, Turtle B.
How to solveBased on the given accelerations:
Turtle C has the smallest acceleration (0.05 cm/s²).Turtle A has the next smallest acceleration (0.1 cm/s²).Turtle B has the largest acceleration (0.2 cm/s²).So the ranking of their accelerations from smallest to largest is: Turtle C, Turtle A, Turtle B.
Based on the given velocities:
Turtle A is going towards the North direction (positive velocity).Turtle B is going towards the South direction (negative velocity).Turtle C is going towards the North direction (positive velocity).As for the change in direction:
Turtle A is accelerating in the same direction as its velocity (North), so it's not changing direction, but its speed is increasing.
Turtle B is accelerating in the opposite direction of its velocity (North), so it will eventually change direction and start moving North when the acceleration overcomes the initial Southward velocity.
Turtle C is accelerating in the same direction as its velocity (North), so it's not changing direction, but its speed is increasing.
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Three turtles are walking in the park. The x-axis is positive towards the North direction. The velocities of the turtles are as follows:
Turtle A has a velocity of 2 cm/s to the North and an acceleration of 0.1 cm/s².
Turtle B has a velocity of 1 cm/s to the South and an acceleration of 0.2 cm/s².
Turtle C has a velocity of 3 cm/s to the North and an acceleration of 0.05 cm/s².
Rank the accelerations of the three turtles from smallest to largest based on their given accelerations.
Are the turtles going towards the North or South direction? Are they changing direction based on their accelerations?
1. In the picture above, how would the amount of kinetic energy in the third pendulum compare to the
amount of potential energy in the first and fourth pendulums?
The kinetic energy would be at its maximum compared to the potential energy which would be less. The potential energy of the first and the last pendulum would be at their maximum
What is kinetic energy?Kinetic energy is the energy an object possesses due to its motion. It is defined as the energy that an object has by virtue of its motion, and is dependent on both its mass and velocity. The formula for calculating kinetic energy is 1/2 times the mass of the object times the square of its velocity, or KE = 1/2 mv^2.
This means that the kinetic energy of an object increases as its mass and velocity increase. For example, a heavy object moving at a high velocity will have a higher kinetic energy than a lighter object moving at a lower velocity. Kinetic energy is a scalar quantity, meaning it has no direction, only magnitude.
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a small metallic sphere has a net charge q1= -2.5 μ c it is held in a stationary position by means of insulating supports. a second small metallic sphere with a net charge q2= -7.8 μ c and a mass of 1.5 g, when the 2 spheres are at a distance of 0.8 m from each other, q2 moves towards q1 with a speed of 22 m/ s. A ¿What is the speed of q2 when the spheres are 0.4 m apart? B. ¿How close does q1 get to q2?
Let's first find the initial electrostatic potential energy of the system. The electrostatic potential energy between two point charges q1 and q2 separated by a distance r is given by:
U = k * q1 * q2 / r
where k is the Coulomb constant, which has a value of approximately 9 x 10^9 N m^2/C^2.
Substituting the given values, we get:
U = (9 x 10^9 N m^2/C^2) * (-2.5 x 10^-6 C) * (-7.8 x 10^-6 C) / 0.8 mU = 2.284 JAt a separation of 0.4 m, the electrostatic potential energy of the system is:
U' = (9 x 10^9 N m^2/C^2) * (-2.5 x 10^-6 C) * (-7.8 x 10^-6 C) / 0.4 m
U' = 9.136 J
The change in potential energy is therefore:
ΔU = U' - U = 6.852 J
This change in potential energy is equal to the kinetic energy of sphere q2, which we can calculate using:
KE = (1/2) * m * v^2
where m is the mass of sphere q2 and v is its velocity.
To find the speed of q2 when the spheres are 0.4 m apart, we can rearrange the above equation and substitute the known values:
v = sqrt(2 * KE / m) = sqrt(2 * ΔU / m) = sqrt(2 * 6.852 J / 0.0015 kg) = 238.9 m/s
Therefore, the speed of sphere q2 when the spheres are 0.4 m apart is approximately 238.9 m/s.
To find how close q1 gets to q2, we can use the conservation of energy principle again. At the closest point of approach, all of the initial potential energy has been converted into kinetic energy, so we can equate the two:
(1/2) * m * v^2 = U
Solving for the separation r, we get:
r = k * q1 * q2 / (2 * KE)
Substituting the known values, we get:
r = (9 x 10^9 N m^2/C^2) * (-2.5 x 10^-6 C) * (-7.8 x 10^-6 C) / (2 * 2.284 J)
r = 0.109 m
Therefore, the closest separation between q1 and q2 is approximately 0.109 m.
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A 2.8 kg block slides along a frictionless surface at 1.1 m/s . A second block, sliding at a faster 4.8 m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.3 m/s.
What was the mass of the second block?
Conservation of momentum is a major law of physics which states that the momentum of a system is constant if no external forces are acting on the system. It is embodied in Newton’s First Law or The Law of Inertia.the mass of the second block is 1.1Kg.
principle of momentum conservation
M1u1 plus M2u2 equals M1v1 and M2V2.
As all collisions were elastic in nature and no energy loss through friction, heat, etc. was taken into account, theoretic calculations alone cannot guarantee that there was a complete transfer of energy.
Consider the scenario where a football with mass M2 is lying on the ground and a bowling ball with mass M1 is hurled at the football at a velocity of
The formula is: (2.8 kg * 1.1 m/s) + (m2 * 4.8 m/s) = (2.3 kg + m2). 2.3 m/s
The formula is 2.8 J + (4.8 m/s m2) = 4.8 J + (2.3 m/s m2).
4.8 m/s m2 = 2.8 J plus (2.3 m/s m2)
4.8 m2 = 2.8 + 2.3 m2
2.3 m2 on each side of the equation
2.5 m = 2.8 m = 2.8 / 2.5\sm = 1.1kg
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Consider a person of mass 59 kg. What would his mass be if he was composed entirely of neutron-star material of density 3 x 10^17 kg/m³? (Assume that his average density is 1000 kg/m³)
Express your answer using two significant figures.
If the human were fully made of neutron star matter, his mass would be roughly 1.8 x 10¹⁶ kg, to two significant numbers.
What elements make up a neutron star?As most protons and electrons will have merged to become neutrons under the extremely dense conditions, neutron stars get their name because they are largely made of neutrons.
Assume the individual has a volume of 59,000 cm3, or about the same as someone with a density of 1000 kg/m3. When we convert this volume to m3, we obtain:
V = 59,000 cm³
V = 0.059 m³
If the individual were made entirely of matter from neutron stars, his mass would be:
m = ρV
m = (3 x 10¹⁷ kg/m³) (0.059 m³)
m = 1.77 x 10¹⁶ kg
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An airplane is flying with a velocity of 240 m/s at an angle of 30.0° with the horizontal, as the drawing shows. When the altitude of the plane is 2.4 km, a flare is released from the plane. The flare hits the target on the ground. What is the angle θ?
Since the flare is released from the plane, it has the same initial velocity as the plane, i.e., 240 m/s at an angle of 30 degrees with the horizontal.
Let's assume that the distance from the plane to the target is d, and the altitude of the plane is h. The angle θ is the angle between the plane and the target.
Using trigonometry, we can write:
tan θ = h/d
We need to find the value of θ. To do that, we need to find the values of h and d.
We know that the altitude of the plane is 2.4 km = 2400 m. Let's call the time it takes for the flare to hit the target t. Since the flare is moving under gravity, its motion can be described as:
h = (1/2)gt^2
where g is the acceleration due to gravity, which is approximately 9.81 m/s^2.
The horizontal distance traveled by the flare in time t is:
d = vt
where v is the horizontal component of the velocity of the flare/plane, which is given by:
v = 240 cos 30° = 240 × √3/2 = 120√3 m/s
Equating the expressions for h and d, we get:
(1/2)gt^2 = vt
Solving for t, we get:
t = 2h/g = 2 × 2400/9.81 = 489.55 s
Substituting this value of t in the expression for d, we get:
d = vt = 120√3 × 489.55 ≈ 70000 m
Now we can find θ:
tan θ = h/d = 2400/70000 ≈ 0.0343
θ = tan^(-1)(0.0343) ≈ 1.96°
Therefore, the angle θ is 1.96 degrees.
A truck accelerates uniformly from rest to 18.5 m/s in 5.7 s along a level stretch of road. Determine the average power (in W) required to accelerate the truck for the following values of the weight (ignore friction). (a)
The average power required to accelerate the truck with a weight of 20,000 N is approximately 116,930.68 W.
The average power required to accelerate an object is given by the formula:
Power = Work / Time
where Work is the change in kinetic energy of the object and Time is the time interval over which the change in kinetic energy occurs.
The change in kinetic energy of the truck is given by:
ΔK = 1/2 * m *[tex]v^2[/tex]
where m is the mass of the truck and v is its final velocity.
Since the truck starts from rest, its initial kinetic energy is zero, so the work done on the truck is equal to its change in kinetic energy. Therefore, the average power required to accelerate the truck is:
Power = Work / Time = ΔK / Time
Substituting the given values, we get:
(a) For weight w = 10,000 N (approximately 1,020 kg):
m = w / g = 10000 N / [tex]9.81 m/s^2[/tex] = 1019.3 kg
ΔK =[tex]1/2 * m * v^2[/tex] = 1/2 * 1019.3 kg * [tex](18.5 m/s)^2[/tex]= 333,036.6 J
Power = ΔK / Time = 333036.6 J / 5.7 s ≈ 58,426.84 W
Therefore, the average power required to accelerate the truck with a weight of 10,000 N is approximately 58,426.84 W.
Note: g is the acceleration due to gravity, which is approximately 9.81 [tex]m/s^2.[/tex]
(b) For weight w = 20,000 N (approximately 2,040 kg):
m = w / g = 20000 N / 9.81 m/s^2 = 2041.2 kg
ΔK = 1/2 * m *[tex]v^2[/tex] = 1/2 * 2041.2 kg * [tex](18.5 m/s)^2[/tex]= 666,073.3 J
Power = ΔK / Time = 666073.3 J / 5.7 s ≈ 116,930.68 W.
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Box ramp
A 1 kg box started from rest downward an
1-meter high ramp inclined at 30 degrees. The
kinetic friction coefficient is 0.3. What was
its speed when it reaches the bottom, in m/s?
A. 3.1
B. 5.5
C. 4.8
D. 1.3
E. 4.4
Answer:
(option A).
Explanation:
The gravitational potential energy of the box at the top of the ramp is given by:
Ep = mgh
where m is the mass of the box, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the ramp (1 m).
Ep = (1 kg)(9.8 m/s^2)(1 m) = 9.8 J
As the box slides down the ramp, its gravitational potential energy is converted into kinetic energy, which is given by:
Ek = (1/2)mv^2
where v is the velocity of the box at the bottom of the ramp.
The work done by the friction force on the box is given by:
W = f * d
where f is the force of friction and d is the distance traveled by the box along the ramp. The force of friction is given by:
f = μmg
where μ is the coefficient of kinetic friction (0.3) and mg is the weight of the box.
f = (0.3)(1 kg)(9.8 m/s^2) = 2.94 N
The distance traveled by the box along the ramp is:
d = h/sin(30°) = 2 m
Therefore, the work done by the friction force is:
W = (2.94 N)(2 m) = 5.88 J
The total mechanical energy of the box at the bottom of the ramp (neglecting air resistance) is equal to the sum of its kinetic energy and the work done by the friction force:
Ek + W = Ep
(1/2)mv^2 + 5.88 J = 9.8 J
Solving for v, we get:
v = sqrt[(2(9.8 J - 5.88 J))/m] = sqrt[(2(3.92 J))/1 kg] = 3.1 m/s
A physics class conducting a research project on projectile motion construct a device that can launch a cricket ball.the launching device is designed so that the ball can be launch at ground level with an initial velocity of 28m/s at an angle of 30 degrees to the horizontal.
Calculate the horizontal of the velocity of the all:
a) initially
B) after 1.0 seconds
C) after 2.0 seconds
A projectile motion is any object thrown into space upon which the only acting force is gravity. The primary force acting on a projectile is gravity.The velocity's horizontal component is 24.25 m/s at time t = 2 seconds.The velocity's horizontal component is 24.25 m/s at time t = 3 seconds.
This doesn’t necessarily mean that other forces do not act on it, just that their effect is minimal compared to gravity.The particle is moving vertically (downwards) along the y-axis due to uniform acceleration.A particle's vertical and horizontal projectile motions can both accelerate: The only force acting on a particle when it is launched into the air at some speed is the acceleration brought on by gravity (g). The downward motion of this acceleration is vertical.
347u = 28 m/s for the starting velocity
projecting at a 30° angle
The horizontal component of velocity is constant since there is no acceleration in the horizontal direction.
Vertical component of speed, u cos = 28 x Cos 30 = 24.25 m/s
The velocity's horizontal component is 24.25 m/s at time t = 2 seconds.
The velocity's horizontal component is 24.25 m/s at time t = 3 seconds
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Please help I need to make sure my answer is correct
The measure that helps your body to stretch is flexibility, so the correct answer is D.
Flexibility for Body Stretching.Flexibility is the ability of your body to move your joints and muscles through their full range of motion. It is an essential component of physical fitness and can help improve posture, balance, and coordination. Regular stretching exercises can increase flexibility and reduce muscle stiffness, which can lead to improved physical performance, decreased risk of injury, and better overall health.
Muscular strength and muscular endurance are related to the ability of your muscles to generate force or sustain effort over time, respectively. While they are important for overall fitness, they are not directly related to flexibility.
In summary, flexibility is the key measure that helps your body to stretch, and regular stretching exercises can improve your flexibility and overall physical fitness.
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An object has greater momentum if it has ?
The more massive or swifter an object is, the more momentum it has.
When a substance has more momentum?A moving thing is more difficult to stop the more momentum it possesses. The amount of momentum an object possesses is influenced by its mass. For instance, while a car driving at the same speed as a baseball can be stopped, it cannot be caught. Because the car is heavier, it has more momentum.
What increases an object's momentum?A moving object's mass and speed both affect its momentum. The greater the object's momentum and the more difficult it is to stop are directly proportional to its weight and speed.
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Suppose an object is weighed with a spring balance, first in air and then whilst totally immersed in water. The readings on the balance are 0.48N and 0.36N respectively. Calculate the density of the object. (2)
The density of the object is is 4000 kg/m³
What is density?Density is the ratio of mass to volume of a body.
To calculate the density of the obeject, we use the formula below
Formula:
D = D'[W/(W-W')]........................ Equation 1Where:
D = Density of the objectD' = Density of waterW = Weight of the object in airW' = Weight of the object in waterFrom the question,
Given:
W = 0.48 NW' = 0.36 ND' = Constant = 1000 kg/m³Susbtitute these values into equation 1
D = 1000[0.48/(0.48-0.36)]D = 1000(0.48/0.12)D = 4000 kg/m³Hence, the density is 4000 kg/m³
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In an inertial system S, an event is observed to take place at point A on the x-axis and 10^-6 s later another event takes place at point B, 900 m further down. Find the magnitude and direction of the velocity of S' with respect to S in which these two events appear simultaneous.
The magnitude of the velocity of S' with respect to S is therefore approximately [tex]9 * 10^(11)[/tex]m/s. then the magnitude is approximately[tex]9 * 10^(11)[/tex] m/s.
the velocity of the system S' with respect to S as v. Since the two events are simultaneous in the system S', they must have the same time coordinate in that system. Let's call this time coordinate t' and the position of the two events in the S' system as (x1', t') and (x2', t').
Using the Lorentz transformation equations, we can relate the coordinates (x1, t1) and (x2, t2) in the S system to the coordinates (x1', t') and (x2', t') in the S' system:
x1' = γ(x1 - vt1)
t' = γ(t1 - vx1/[tex]c^2[/tex])
x2' = γ(x2 - vt2)
t' = γ(t2 - [tex]vx2/c^2[/tex])
where γ = 1/√(1 - [tex]v^2/c^2[/tex]) is the Lorentz factor.
Since the two events are separated by 900 m and a time interval of 10^-6 s, we have:
x2 - x1 = 900 m
t2 - t1 = [tex]10^-6[/tex] s
Using the above equations, we can solve for v. First, we can eliminate t' by equating the two expressions for t':
γ(t1 - vx1/[tex]c^2[/tex]) = γ(t2 - vx2/[tex]c^2[/tex])
Simplifying this expression, we get:
v =[tex]c^2[/tex](x2 - x1)/(t2 - t1)(x1 + x2)
Plugging in the given values, we get:
v = c^2(900 m)/([tex]10^-6[/tex] s)(0 m + 900 m) = 9 × [tex]10^11[/tex] m/s
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Two friends, Burt and Ernie, are standing at opposite ends of a uniform log that is floating in a lake. The log is 4.0 m
long and has mass 250 kg
. Burt has mass 30.0 kg
and Ernie has mass 39.0 kg
. Initially the log and the two friends are at rest relative to the shore. Burt then offers Ernie a cookie, and Ernie walks to Burt's end of the log to get it.
a) Relative to the shore, what distance has the log moved by the time Ernie reaches Burt? Neglect any horizontal force that the water exerts on the log and assume that neither Burt nor Ernie falls off the log.
The log moves a distance of 2.0 m relative to the shore when Ernie reaches Burt.
Since the log is uniform, we can treat it as a system of three point masses: one at each end representing Burt and Ernie, and one at the center representing the center of mass of the log. We can use conservation of momentum to solve this problem.
Initially, the momentum of the system is zero since everything is at rest. When Ernie walks to Burt's end of the log, he exerts a force on the log, which in turn exerts an equal and opposite force on Ernie. This force is internal to the system and does not change the momentum of the system. Therefore, the momentum of the system is still zero after Ernie reaches Burt.
Total momentum = 0 (since the center of mass is at rest)
Burt's momentum = 0 (since he is at rest)
Ernie's momentum = 0 (since he is at rest)
After Ernie walks:
Total momentum = 0 (since the center of mass remains at rest)
Burt's momentum = 0 (since he remains at rest)
Ernie's momentum = (39.0 kg) * v, where v is the velocity of Ernie and the log in the opposite direction
Since the total momentum is conserved, we can equate the momentum before and after Ernie walks:
0 = (39.0 kg) * v
v = 0 m/s
the distance that the log moves when Ernie reaches Burt is simply the distance between the initial and final positions of Ernie, which is half the length of the log:
distance = (1/2) * (4.0 m) = 2.0 m
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Two pure tones Cs and Gs, with frequencies from the Pythagorean diatonic scale, are sounded simultaneously. Find
a) the frequencies of the three combination tones and
b) the notes on the Pythagorean scale to which these tones belong.
The combination tones correspond to the notes Bb, D, and Cs on the Pythagorean diatonic scale.
What is Frequency?
Frequency is the number of occurrences of a repeating event per unit of time. In other words, it is the rate at which a wave oscillates or completes one cycle. The unit of frequency is hertz (Hz), which is equivalent to one cycle per second.
When two pure tones with frequencies f1 and f2 are sounded simultaneously, several additional frequencies, known as combination tones, can be produced. The three most important combination tones are:
The sum tone, which has a frequency equal to the sum of the two original frequencies: f1 + f2
The difference tone, which has a frequency equal to the difference between the two original frequencies: |f1 - f2|
The octave tone, which has a frequency twice that of the lower of the two original frequencies: 2f1 or 2f2
In this case, we have two pure tones Cs and Gs with frequencies from the Pythagorean diatonic scale. We need to first determine the frequencies of these two tones. According to the Pythagorean tuning system, the frequency ratios for Cs and Gs are:
Cs:G = 9:8
Cs:fundamental = 2:1 (assuming Cs is one octave above the fundamental)
Gs:fundamental = 3:1 (assuming Gs is one octave and a fifth above the fundamental)
Let's assume that the fundamental frequency is f0. Then we can write:
Cs = 2f0 * (9/8) = 9f0/4
Gs = 4f0 * (3/2) * (9/8) = 27f0/8
a) To find the combination tones, we need to apply the equations above. The sum tone has a frequency of:
f1 + f2 = Cs + Gs = (9f0/4) + (27f0/8) = 45f0/8
The difference tone has a frequency of:
|f1 - f2| = |Cs - Gs| = |(9f0/4) - (27f0/8)| = 9f0/8
The octave tone has a frequency of:
2f1 = 2Cs = 9f0/2 = 9f0
Therefore, the three combination tones have frequencies of 45f0/8, 9f0/8, and 9f0.
b) To determine the notes on the Pythagorean scale to which these tones belong, we need to find the closest notes on the scale to each of the combination tones. The Pythagorean scale is based on a series of perfect fifths, so we can use the frequency ratios of 3:2 to determine the frequency of each note relative to the fundamental frequency f0.
The closest notes on the Pythagorean scale to the combination tones are:
45f0/8 is closest to the note Bb, which has a frequency of 3f0/2
9f0/8 is closest to the note D, which has a frequency of 9f0/8
9f0 is closest to the note Cs, which has a frequency of 9f0/4
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