The required coordinates of the 4.37 kg particle is (-0.415 m, -0.138 m) where the center of mass of this three particle system has the coordinates (-0.666 m, -0.381 m).
4.20 kg particle coordinates = (-1.92 m, 0.878 m)
2.04 kg particle coordinates = (0.563 m, -0.310 m)
Center of mass coordinates = (-0.666 m, -0.381 m)
We need to find the coordinates of 4.37 kg particle(a) x coordinate
If the x-coordinate of the center of mass is -0.666 m, then for the three-particle system, we can say:
4.20x1 + 2.04x2 + 4.37x3 = 3 × (-0.666)4.20(-1.92) + 2.04(0.563) + 4.37x3 = -1.998x3 = (4.20)(-1.92) + (2.04)(0.563) - 3(-0.666) / 4.37x3 = -0.415m
(b) y coordinate
If the y-coordinate of the center of mass is -0.381 m, then for the three-particle system, we can say:
4.20y1 + 2.04y2 + 4.37y3 = 3 × (-0.381)4.20(0.878) + 2.04(-0.310) + 4.37y3
= -1.143y3 = (4.20)(0.878) + (2.04)(-0.310) - 3(-0.381) / 4.37y3
= -0.138m
Therefore, the coordinates of the 4.37 kg particle is (-0.415 m, -0.138 m).
Hence, the required coordinates of the 4.37 kg particle is (-0.415 m, -0.138 m) where the center of mass of this three particle system has the coordinates (-0.666 m, -0.381 m).
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The (a) x and (b) y coordinates of the third particle so that it's center of mass has the coordinates (-0.666 m, -0.381 m) are (-0.087 m, -0.799 m), respectively.
Two particles A 4.20 kg particle with xy coordinates (-1.92 m, 0.878 m). 2.04 kg particle with xy coordinates (0.563 m, -0.310 m)
Third particle 4.37 kg.
The center of mass of the three-particle system has the coordinates (-0.666 m, -0.381 m)
Center of mass: It is the point where the system of particles behaves as if the entire mass is concentrated at this point.
Let the x and y coordinates of the third particle be x3 and y3, respectively.
[tex]x_{cm}=\frac{\sum_{i} m_{i} x_{i}}{\sum_{i} m_{i}}[/tex]
And, the y-coordinate of the center of mass is given as:
[tex]y_{cm}=\frac{\sum_{i} m_{i} y_{i}}{\sum_{i} m_{i}}[/tex]
Let’s consider the x-coordinate first.The sum of masses of all three particles is given as: 4.20 kg + 2.04 kg + 4.37 kg = 10.61 kg
The sum of masses of the first two particles is given as:
4.20 kg + 2.04 kg = 6.24 kg
Hence, the mass of the third particle is: 10.61 kg - 6.24 kg = 4.37 kg
Now, let's calculate the x-coordinate of the third particle using the center of mass formula:
[tex]x_{cm}=\frac{\sum_{i} m_{i} x_{i}}{\sum_{i} m_{i}}[/tex]
[tex]x_{cm}=\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}[/tex]
Here, [tex]m_1=4.20 \ kg,[/tex]
[tex]x_1=-1.92 \ m (coordinates of first particle) m_2=2.04 \ kg,[/tex]
[tex]x_2=0.563 \ m (coordinates of second particle) m_3=4.37 \ kg,[/tex]
[tex]x_3=??[/tex] (coordinates of third particle) and the center of mass is at [tex]x_{cm}=-0.666 \ m[/tex]
[tex]-0.666 \ m=\frac{(4.20 \ kg)(-1.92 \ m)+(2.04 \ kg)(0.563 \ m)+(4.37 \ kg)(x_3)}{10.61 \ kg}[/tex]
Solving for
[tex]x_3:x_3=-0.087 \ m[/tex]
Now, let's calculate the y-coordinate of the third particle using the center of mass formula:
[tex]y_{cm}=\frac{\sum_{i} m_{i} y_{i}}{\sum_{i} m_{i}}[/tex]
[tex]y_{cm}=\frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}[/tex]
Here, m_1=4.20 \ kg,
[tex]y_1=0.878 \ m (coordinates of first particle) m_2=2.04 \ kg,[/tex]
[tex]y_2=-0.310 \ m (coordinates of second particle) m_3=4.37 \ kg, y_3=??[/tex] (coordinates of third particle) and the center of mass is at [tex]y_{cm}=-0.381 \ m[/tex]
[tex]-0.381 m = [(4.20 kg)(0.878 m) + (2.04 kg)(-0.310 m) + (4.37 kg)(y3)] / 10.61 kg[/tex]
Solving for [tex]y_3:[/tex]
y_3=-0.799 \ m
Therefore, the (a) x and (b) y coordinates of the third particle are (-0.087 m, -0.799 m), respectively.
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A 2,500 Hz sound wave travels with a speed of 15 m/s in water. A paleontologist measures
the valley to the second valley of the wave to be 7.5 cm.
➤What is the (a) period? What is the (b) frequency? What is the (c) wavelength?
The answers are A. The period of the wave is 4 × 10⁻⁴ s, B. The frequency is 2500 Hz and C. The wavelength is 6 cm.
A sound wave is a type of wave that travels through the medium by compressing and expanding the particles of the medium. These waves have certain characteristics that are used to measure their properties. The following are the answers to the given question: A 2,500 Hz sound wave travels with a speed of 15 m/s in water. A paleontologist measures the valley to the second valley of the wave to be 7.5 cm.a) The period of a wave is the time it takes to complete one cycle. The formula for calculating the period of a wave is Period = 1/Frequency. Here, the frequency of the wave is 2500 Hz. Hence, the period of the wave can be calculated as Period = 1/2500 Hz = 4 × 10⁻⁴ s.b) The frequency of a wave is the number of cycles that pass a point in one second. The formula for calculating the frequency of a wave is Frequency = 1/Period. Here, the period of the wave is 4 × 10⁻⁴ s. Hence, the frequency of the wave can be calculated as Frequency = 1/4 × 10⁻⁴ s = 2500 Hz.c) The wavelength of a wave is the distance between two successive points on the wave that are in phase. The formula for calculating the wavelength of a wave is Wavelength = Wave speed / Frequency. Here, the wave speed of the sound wave is 15 m/s and the frequency of the wave is 2500 Hz. Hence, the wavelength of the wave can be calculated as Wavelength = 15 / 2500 = 0.006 m = 6 cm.For more questions on frequency
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you jumped from the same height into water as opposed to onto concrete, the impulse required to reduce your speed to zero velocity, assuming you have same posture at impact in cach case, would be the same True / False
False. The impulse required to reduce your speed to zero velocity would be different when jumping into water compared to jumping onto concrete.
The impulse required to reduce your speed to zero velocity would not be the same when jumping into water compared to jumping onto concrete. The impulse is equal to the change in momentum, which is determined by the mass and velocity of the object.
When jumping into water, the water exerts a greater resistive force compared to the concrete, which results in a longer deceleration time and a smaller impulse. The water acts as a cushion, spreading out the force over a longer duration.
On the other hand, when jumping onto concrete, the deceleration time is shorter, resulting in a larger impulse and potentially higher impact forces. The concrete does not provide the same cushioning effect as water.
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Monochromatic light from a distant source is incident on a slit 0.755 mm wide. On a screen 1.98 m away, the distance from the central maximum of the diffraction pattern to he first minimum is measured to be 1.35 mm For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Single-slit diffraction.
Calculate the wavelength of the light. Express your answer in meters.
The wavelength of the Monochromatic light is 5.17 × 10⁻⁷ m.
A narrow opening with a width of 0.755 mm is illuminated by monochromatic light originating from a distant source.
At a distance of 1.98 m from the narrow opening, the distance between the central maximum and the first minimum of the diffraction pattern is found to be 1.35 mm.
The wavelength of the light needs to be calculated. We know that the central maximum is formed at the center of the diffraction pattern. The equation provided allows us to determine the distance between the central maximum and the first minimum.
[tex]$$D_m = \frac{m\lambda L}{a}$$[/tex]
where m = 1, a = 0.755 mm, L = 1.98 m, and [tex]$D_m$[/tex] = 1.35 mm.
After plugging in the given values into the equation mentioned above, we obtain the following result.
[tex]$$1.35 \times 10^{-3} = \frac{(1)(\lambda)(1.98)}{0.755 \times 10^{-3}}$$[/tex]
[tex]$$\lambda = \frac{1.35 \times 10^{-3} \times 0.755 \times 10^{-3}}{1.98} = 5.17 \times 10^{-7}m$$[/tex]
Hence, the wavelength of the light is 5.17 × 10⁻⁷ m.
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An EM wave has an electric field given by E = (200 V/m) [sin ((0.5m-¹)-(5 x 10°rad/s)t)] 3. Find a) Find the wavelength of the wave. b) Find the frequency of the wave qool A (3q 1) # c) Write down the corresponding function for the magnetic field.
The corresponding function for the magnetic field is B = 6.67 x 10⁻⁷ [sin ((0.5m⁻¹)-(5 x 10⁹ rad/s)t)] T.
a) Calculation of the wavelength of the waveThe equation for wavelength is given by λ = 2π/k, where k is the wavenumber.We can find k from the equation k = 2π/λSubstituting the value of λ, we get:k = 2π/0.5m⁻¹k = 12.56 m⁻¹Therefore,λ = 2π/kλ = 0.5 m b) Calculation of frequency of the waveFrequency (ν) is given by the equation ν = ω/2πSubstituting the values of ω, we getν = 5 x 10¹⁰ rad/s / 2πν = 7.96 x 10⁹ Hz c) Expression for the magnetic fieldThe equation for the magnetic field (B) is given by B = E/c, where c is the speed of light.Substituting the values of E and c, we get:B = (200 V/m) [sin ((0.5m⁻¹)-(5 x 10⁹ rad/s)t)] / 3 x 10⁸ m/sB = 6.67 x 10⁻⁷ [sin ((0.5m⁻¹)-(5 x 10⁹ rad/s)t)] TTherefore, the corresponding function for the magnetic field is B = 6.67 x 10⁻⁷ [sin ((0.5m⁻¹)-(5 x 10⁹ rad/s)t)] T.
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An electron is flying through space and traverses a volume containing protons. However, no X ray is produced. Why? The proton desinty in space is very low so close encounters are rare. Physics works differently in other parts of the universe. The X ray is shifted to a longer wave length. none of these is correct.
The correct answer is option a) "The proton density in space is very low so close encounters are rare."
The lack of X-ray production by the electron can be attributed to the low density of protons in space, making close encounters between the electron and protons rare. X-rays are typically generated when high-energy electrons interact with matter, causing the electrons to decelerate rapidly and emit photons in the X-ray range. In this scenario, however, the scarcity of protons in the volume through which the electron is passing inhibits significant interactions.
Option b, suggesting that physics works differently in other parts of the universe, is not a plausible explanation in this context. The fundamental laws of physics, including the behavior of electrons and photons, remain consistent throughout the universe. Therefore, it is not a valid reason for the absence of X-ray production in this particular situation.
Option c proposes that the X-ray is shifted to a longer wavelength. However, this is not applicable because the absence of X-ray production cannot be attributed to a change in the wavelength of the emitted X-rays. Rather, it is primarily due to the low proton density.
Therefore, the correct answer is option a, as it accurately explains the lack of X-ray production by the electron passing through the volume with protons. The rare encounters between the electron and the low-density protons in space hinder the generation of X-rays.
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A uniform plane Electromagnetic wave is expressed by E (2,t) = 1600 cos (10?mt - Bz)a, v/m and Hz :) = 4.8 cos (10?mt - Bz), A/m. The wave propagates in a perfect dielectric along the z-axis with propagation velocity of v = 2 x 108 m/s. Find the following: (a) the phase constant, B (4 marks) (b) the wavelength, A ( 4 marks) (c) the intrinsic impedance, n (4 marks)
Given: An electromagnetic wave is expressed by E(2,t) = 1600 cos (10πmt - Bz)a, V/m and Hz := 4.8 cos (10πmt - Bz), A/m.
The wave propagates in a perfect dielectric along the z-axis with a propagation velocity of v = 2 × 108 m/s.
The equation of the electromagnetic wave is given as:
E(z, t) = 1600 cos(10πmt − Bz) a
The wave travels along the z-direction, so its phase is given by:
Bz=2π/λ z, where λ is the wavelength.
The phase constant can be determined as:
B = 2π/λ = 10π m
Since the wave propagates in a perfect dielectric medium, the intrinsic impedance of the medium is given by:
μ0/ε0where μ0 and ε0 are the permeability and permittivity of free space, respectively.
Intrinsic Impedance (η) = √(μ0/ε0) = 377 Ω
Thus, the intrinsic impedance is 377 Ω.
An electromagnetic wave is expressed by E(2,t) = 1600 cos (10πmt - Bz)a, V/m and Hz := 4.8 cos (10πmt - Bz), A/m. The wave propagates in a perfect dielectric along the z-axis with a propagation velocity of v = 2 × 108 m/s.
Therefore, the wavelength can be calculated as:
A = v/f = v/λ where v is the velocity of propagation, f is the frequency, and λ is the wavelength.
f = 10 MHz = 10 × 106 Hz, λ = v/f = 2 × 108/10 × 106= 20 m
Hence, the wavelength is 20 m.
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A solenoid with 465 turns has a length of 6.50 cm and a cross-sectional area of 2.60 x 10⁻⁹ m². Find the solenoid's inductance and the average emf around the solenoid if the current changes from +3.50 A to -3.50 A in 6.83 x 10⁻³ s. (a) the solenoid's inductance (in H) _____ H (b) the average emf around the solenoid (in V)
_____ V
A solenoid with 465 turns has a length of 6.50 cm and a cross-sectional area of 2.60 x 10⁻⁹ m².
The expression for the inductance of the solenoid is given by the formula:
L = (μ_0*N^2*A)/L where L = length of the solenoid N = number of turns A = cross-sectional area m = permeability of free space μ_0 = 4π x 10⁻⁷ H/m
∴ Substituting the given values in the above formula,
L = (μ_0*N^2*A)/L= (4π x 10⁻⁷ x 465² x 2.60 x 10⁻⁹)/0.065L = 8.14 x 10⁻³ H
The average emf around the solenoid (in V)
The emf around the solenoid is given by the formula:
emf = -L((ΔI)/(Δt)) where emf = electromotive force L = inductance ΔI = change in current Δt = change in time
∴Substituting the given values in the above formula, emf = -L((ΔI)/(Δt))= -8.14 x 10⁻³(((-3.50 A) - (3.50 A))/(6.83 x 10⁻³ s))= 1.65 V
Thus, The solenoid's inductance = 8.14 x 10⁻³ H.
The average emf around the solenoid = 1.65 V.
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It is proposed to work with a heating system, through extracting groundwater at 50 Fahrenheit to heat a house up to 70 Fahrenheit. Groundwater drops by 12 degrees Fahrenheit. The house demands 75,000Btu/h.
Calculate the minimum flow in lbm/h of water needed to complete this task.
Enter only the numerical value
The minimum flow rate of water needed to complete this heating task is approximately 6250 lbm/h.
To calculate the minimum flow rate of water needed to complete this heating task, we need to consider the energy balance equation:
Flow rate (lbm/h) * Specific heat capacity of water (Btu/lbm°F) * Temperature drop (°F) = Heating demand (Btu/h)
Given:
Groundwater temperature in = 50 °F
Heating target temperature out = 70 °F
Temperature drop = 12 °F
Heating demand = 75,000 Btu/h
Let's calculate the minimum flow rate of water:
Flow rate * Specific heat capacity of water * Temperature drop = Heating demand
Flow rate * (1 Btu/lbm°F) * 12 °F = 75,000 Btu/h
Flow rate = 75,000 Btu/h / (1 Btu/lbm°F * 12 °F)
Flow rate = 75,000 lbm/h / 12
Flow rate ≈ 6250 lbm/h
Therefore, the minimum flow rate of water needed to complete this heating task is approximately 6250 lbm/h.
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In a football stadium, an announcer call plays over a loud speaker. The sound wave has a frequency of 192 Hz. If it take 3.13 seconds to reach a fan that is seated 89.68 m away from the loud speaker, find the speed of the sound wave? Answer to the hundreths place or two decimal places.
The speed of the sound wave in the football stadium is approximately 288.43 m/s.
To find the speed of the sound wave, we can use the formula: speed = distance / time. Given that the distance from the loud speaker to the fan is 89.68 m and it takes 3.13 seconds for the sound wave to reach the fan, we can substitute these values into the formula. Therefore, the speed of the sound wave is 89.68 m / 3.13 s = 28.64 m/s.
However, this calculation only provides the speed of the sound wave over that specific distance. To obtain the actual speed of the sound wave, we need to consider the frequency of the wave.
The formula for the speed of a sound wave is speed = frequency × wavelength. Since we know the frequency of the sound wave is 192 Hz, we need to calculate the wavelength.
The wavelength of a sound wave can be determined using the formula wavelength = speed / frequency. Plugging in the previously calculated speed (28.64 m/s) and the frequency (192 Hz), we can find the wavelength: wavelength = 28.64 m/s / 192 Hz = 0.1492 m.
Now, we can use the calculated wavelength to find the actual speed of the sound wave using the formula speed = frequency × wavelength. With the frequency of 192 Hz and the wavelength of 0.1492 m, the speed of the sound wave is 192 Hz × 0.1492 m = 28.71 m/s.
Therefore, the speed of the sound wave in the football stadium is approximately 28.71 m/s, rounded to two decimal places, or 288.43 m/s.
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A load of +9 nC is placed on the x axis at x = 3.2 m, and a load of -25 nC is placed at x = -5.9 m. What is the magnitude of the electric field at the origin?From your response to a decimal place.
The magnitude of the electric field at the origin is 29.44 N/C (to two decimal places).
The given data are;
A load of +9 nC is placed on the x-axis at x = 3.2 mA load of -25 nC is placed at x = -5.9 m. The objective is to calculate the magnitude of the electric field at the origin. Now we will use the formula below;
E=k∑(q÷r²) Where k is the Coulomb constant
k = 9 × 10⁹ N.m²/C²q is the magnitude of the point charge in Coulombs (C)r is the distance between the point charge and the field position.
The electric field is a vector quantity with a magnitude given by
E = F/q where F is the force experienced by a unit charge (+1 C) placed at that point.
The electric field is a vector quantity. Its direction is the same as the direction of the force experienced by a positive test charge (+1 C) placed at that point by the other charges.
q=9 nC= 9 × 10⁻⁹ C
x=3.2 m
Distance between point charge and origin (r)=3.2 m
∴ E₁=k(q₁/r₁²)=9×10⁹×(9×10⁻⁹)/3.2²=71.484375 N/C
According to the principle of superposition, we can add the electric fields produced by each charge at the origin to obtain the net electric field.
Distance between point charge and origin (r)=5.9 m
∴ E₂=k(q₂/r₂²)=9×10⁹×(-25×10⁻⁹)/5.9²=-100.9290391 N/C
According to the principle of superposition,
the net electric field at the origin= E₁ + E₂=71.484375-100.9290391=-29.4446639 N/C
Therefore, the magnitude of the electric field at the origin is 29.44 N/C (to two decimal places).
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A light ray is incident at an angle of 20° on the surface between air and water. At what angle in degrees does the refracted ray make with the perpendicular to the surface when is incident from the air side? Use index of refraction for air as 1.0 while water 1.33. (Express your answer in 2 decimal place/s,
When a light ray is incident at an angle of 20° on the surface between air and water, the refracted ray makes an angle of approximately 14.68° with the perpendicular to the surface when it is incident from the air side.
The angle between the incident ray and the perpendicular to the surface is known as the angle of incidence. In this case, the angle of incidence is 20°. The angle between the refracted ray and the perpendicular to the surface is known as the angle of refraction.
To find the angle of refraction, we can use Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the indices of refraction of the two media involved.
Given that the index of refraction for air is 1.0 and for water is 1.33, we can set up the following equation:
sin(20°) / sin(angle of refraction) = 1.0 / 1.33
Rearranging the equation and solving for the angle of refraction, we find:
sin(angle of refraction) = sin(20°) * 1.33 / 1.0
angle of refraction ≈ arcsin(sin(20°) * 1.33 / 1.0)
Using a calculator, we find that the angle of refraction is approximately 14.68°. Therefore, the refracted ray makes an angle of approximately 14.68° with the perpendicular to the surface when it is incident from the air side.
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While drilling a well a rock layer is encountered at 8300ft. depth with an excess pressure (overpressure) of 150 psi. An overpressure zone has fluid pressures in excess of the hydrostatic gradient. If the overburden density is 2500 kg/m^3 and the fluid column is water what is the effective stress at this depth?
The effective stress at a depth of 8300 ft with an overpressure of 150 psi, an overburden density of 2500 kg/m³, and a fluid column of water is 5.29 MPa.
Given:
Overpressure = 150 psi
Depth of rock layer = 8300 ft
Overburden density = 2500 kg/m³
Fluid column = Water
Formula used:
Effective stress = Overburden pressure - Fluid pressure
At a depth of 8300 ft, the overburden pressure can be calculated as:
P = γ x d
Where,
γ = Overburden density = 2500 kg/m³
d = Depth of rock layer in meters (convert from ft to m) = 8300 ft x 0.3048 m/ft = 2529.84 m
Substituting the values:
P = 2500 kg/m³ x 2529.84 m
P = 6,324,600 Pa
The fluid pressure can be converted from psi to Pa by multiplying it with 6894.75:
Fluid pressure = 150 psi x 6894.75 = 1,034,212.5 Pa
Therefore, the effective stress at this depth will be:
Effective stress = Overburden pressure - Fluid pressure
= 6,324,600 Pa - 1,034,212.5 Pa
= 5,290,387.5 Pa
= 5.29 MPa
Hence, the effective stress at this depth is 5.29 MPa.
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A car starts from rest and accelerates with a constant acceleration of 2 m/s for 3 s. The car continues for 5 s at constant velocity. The driver then applied the brakes and the car stopped after it raveled 50 m (from the point when the brakes applied and the stopping point). Calculate the average velocity of the car for the entire trip.
The average velocity of the car for the entire trip is approximately -3.44 m/s. The negative sign indicates that the car's motion is in the opposite direction of its initial motion.
The average velocity of the car for the entire trip can be calculated by dividing the total displacement by the total time. The car accelerates for 3 s, travels at constant velocity for 5 s, and then decelerates to a stop over a distance of 50 m. By calculating the displacements and times for each segment of the trip, we can determine the average velocity.
First, let's calculate the displacement and time for each segment of the trip. During the acceleration phase, the car starts from rest and accelerates with a constant acceleration of 2 m/s² for 3 seconds. Using the kinematic equation, we can find the displacement during this phase: d1 = (1/2) * a * t² = (1/2) * 2 * (3²) = 9 m.
During the constant velocity phase, the car travels for 5 seconds at a constant velocity, so the displacement during this phase is d2 = v * t = 2 m/s * 5 s = 10 m.
Finally, during the deceleration phase, the car stops after traveling 50 m. The displacement during this phase is d3 = -50 m (negative because it is in the opposite direction of the car's initial motion).
Now, we can calculate the total displacement: total displacement = d1 + d2 + d3 = 9 m + 10 m - 50 m = -31 m.
The total time for the entire trip is 3 s (acceleration) + 5 s (constant velocity) + time to stop. Since the car stops after traveling 50 m, we can calculate the time to stop using the equation v² = u² + 2ad, where u is the initial velocity (2 m/s), a is the deceleration (assumed to be the same as the acceleration, -2 m/s²), and d is the displacement (-50 m). Solving for time, we find time to stop = (v - u) / a = (0 - 2) / -2 = 1 s. Therefore, the total time is 3 s + 5 s + 1 s = 9 s.
Finally, we can calculate the average velocity by dividing the total displacement by the total time: average velocity = total displacement / total time = -31 m / 9 s ≈ -3.44 m/s.
Therefore, the average velocity of the car for the entire trip is approximately -3.44 m/s. The negative sign indicates that the car's motion is in the opposite direction of its initial motion.
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A 48-kg person and a 75-kg person are sitting on a bench 0.80 m close to each other. Calculate the magnitude of the gravitational force each exerts on the other. (Hint: G = 6.67x10^-11 N-m^2/kg^2)
The magnitude of the gravitational force each person exerts on the other is 1.49 x 10^-8 N.
Newton's Law of Universal GravitationThe force of gravity (F) between two objects is directly proportional to the product of their masses (m1 and m2) and inversely proportional to the square of the distance (r) between them. The formula for the gravitational force between two masses is:F = G * (m1 * m2) / r²
where G is the gravitational constant (6.67 x 10^-11 N m²/kg²).
Given information: Mass of person 1 (m1) = 48 kg, Mass of person 2 (m2) = 75 kg, distance (r) = 0.8 m.
To calculate the force of gravity (F) between the two people, we can use the above formula:
F = G * (m1 * m2) / r²
F = 6.67 x 10^-11 N m²/kg² * ((48 kg) * (75 kg)) / (0.8 m)²
F = 6.67 x 10^-11 N m²/kg² * (3600 kg²) / (0.64 m²)
F = 1.49 x 10^-8 N
The magnitude of the gravitational force each person exerts on the other is 1.49 x 10^-8 N. It should be noted that the force of gravity is an attractive force, meaning that each person attracts the other. Therefore, both people would experience the same force.
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A storage tank at STP contains 28.9 kg of nitrogen (N2) What is the volume of the tank? What is the pressure if an additional 28.1 kg of nitrogen is added without changing the temperature?
The volume of the tank is approximately 24046.31 liters.
The pressure in the tank, after adding the additional nitrogen, is approximately 22.963 atm.
We'll use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.
Finding the volume of the tank:
At STP (Standard Temperature and Pressure), the temperature is 273.15 K, and the pressure is 1 atm. We need to find the volume of the tank when it contains 28.9 kg of nitrogen (N2).
First, we need to determine the number of moles of nitrogen using its molar mass (M):
M(N2) = 28.02 g/mol
Number of moles (n) = mass / molar mass
n = 28.9 kg / (28.02 g/mol) = 1030.532 mol
Now, let's calculate the volume (V) using the ideal gas law:
PV = nRT
V = nRT / P
V = (1030.532 mol) * (0.0821 L·atm/mol·K) * (273.15 K) / (1 atm)
V ≈ 24046.31 L
Finding the pressure after adding more nitrogen:
Now, let's calculate the pressure when an additional 28.1 kg of nitrogen is added to the tank, without changing the temperature.
First, we need to determine the total number of moles of nitrogen:
Total moles = initial moles + additional moles
Total moles = 1030.532 mol + (28.1 kg / (28.02 g/mol))
Total moles ≈ 1030.532 mol + 1000 mol ≈ 2030.532 mol
Now, let's calculate the pressure (P) using the ideal gas law:
PV = nRT
P = nRT / V
P = (2030.532 mol) * (0.0821 L·atm/mol·K) * (273.15 K) / (24046.31 L)
P ≈ 22.963 atm
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The record of the Kobe earthquake is measured using an accelerometer. Use the program you wrote in Problem to compute the amplitude spectrum of the Kobe earthquake data and discuss what frequencies are dominant. You will need to plot the time domain data and the frequency domain data (the amplitude spectrum) out. Note that the data file has two columns: the first column is time and the second column is acceleration..
The amplitude spectrum of the Kobe earthquake data can be used to determine the dominant frequencies present in the data. By analyzing the highest amplitude in the spectrum, we can identify the frequency components that are most prominent in the earthquake data.
The record of the Kobe earthquake was measured using an accelerometer. The program previously written in Problem can be utilized to calculate the amplitude spectrum of the Kobe earthquake data. In order to plot the data in the time domain and frequency domain (the amplitude spectrum), the data file with two columns - time and acceleration - needs to be considered. Initially, it is important to create a graph of acceleration versus time. Subsequently, the FFT function is applied to obtain the frequency-domain data. When plotting the frequency domain data, it is crucial to understand that the frequency axis represents the number of cycles of the periodic waveform per second, which is expressed in Hertz (Hz).
The frequencies that are prominent in the Kobe earthquake data can be determined by analyzing the amplitude spectrum. An amplitude spectrum illustrates the amplitudes of different frequency components present in a signal. The highest amplitude in the amplitude spectrum signifies the dominant frequency, representing the natural frequency of the system being observed. In simpler terms, the dominant frequency is the frequency at which the system oscillates most intensely.
Hence, by examining the amplitude spectrum of the Kobe earthquake data, we can identify the frequency components that are prominent in the data, as indicated by the highest amplitude.
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A 1581.64 kg tank runs out of brakes when it achieves a speed of 34.83 mi/hr. What linear momentum will you be experiencing?
Remember to perform the necessary conversions before solving.
Express your answer WITHOUT DECIMALS.
Florence, mass 55 kg, is running the 100 m dash at a track and field meet. During her sprint, she uses 5300 J of energy, daya is 86% efficient at converting her energy into kinetic energy. What is her final velocity? [13]
Answer: The final velocity of Florence is 13.89 m/s.
Mass of Florence, m = 55 kg
Distance covered by Florence = 100 m
Efficiency of her sprint = 86 % = 0.86
Energy used by Florence = 5300 J
Let's derive the formula for kinetic energy and solve for final velocity.
Final Kinetic energy, K = 0.5 mv²
where, K = Kinetic energy of the body m = mass of the body, v = final velocity of the body. Using work-energy theorem, we know that the work done on a body is equal to its change in kinetic energy. The equation for work done on a body, W is given by
W = K - Ki
where, Ki is the initial kinetic energy of the body.
In this case, initial kinetic energy is 0 as Florence was initially at rest. Work done is given by the energy used by her.
Hence, we can rewrite the equation as 5300 J = K - 0
Substituting the formula for K, we get
5300 = 0.5 * 55 * v²
v² = 5300 / 27.5
v² = 192.7273
Taking the square root of both sides, we get v = 13.89 m/s. Therefore, the final velocity of Florence is 13.89 m/s.
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A motorcycle is traveling at 25 m/s when the rider notices a traffic jam way ahead of them in the distance. Assuming the motorcyclist starts braking with an acceleration of -5 m/s^2 instantly upon noticing the traffic jam, how long (in seconds) does it take the rider to come to a complete stop? (Your answer should be in units of seconds, but just write the number part of your answer.)
The rider takes 5 seconds for the motorcyclist to come to a complete stop. The time it takes for the motorcyclist to come to a complete stop, we can use the kinematic equation that relates velocity, acceleration, and time:
v = u + at
v is the final velocity (0 m/s since the motorcyclist comes to a complete stop),
u is the initial velocity (25 m/s),
a is the acceleration (-5 m/s²),
t is the time we need to find.
t = (v - u) / a
Substituting the given values into the equation:
t = (0 - 25) / (-5)
Simplifying the expression:
t = 25 / 5
t = 5 seconds
Therefore, it takes the motorcyclist 5 seconds to come to a complete stop.
The time it takes for an object to come to a stop can be determined using the kinematic equation that relates velocity, acceleration, and time. In this case, the initial velocity of the motorcyclist is 25 m/s, and the acceleration is -5 m/s² (negative since it is deceleration or braking).
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Please explain how the response of Type I superconductors differ from that of Type Il superconductors when an external magnetic field is applied to them. What is the mechanism behind the formation of Cooper pairs in a superconductor? To answer this question, you can also draw a cartoon or a diagram if it helps, by giving a simple explanation in your own words
Cooper pairs have a net charge of 2e (twice the elementary charge) and behave as bosons rather than fermions. Due to their bosonic nature, Cooper pairs can condense into a collective quantum state, known as the superconducting state, with remarkable properties such as zero electrical resistance and perfect diamagnetism.
Type I and Type II superconductors exhibit different responses to an external magnetic field.
Type I superconductors:
Type I superconductors have a single critical magnetic field (Hc) below which they exhibit perfect diamagnetic behavior, expelling all magnetic field lines from their interior.
When the applied magnetic field exceeds the critical field, the superconductor undergoes a phase transition and loses its superconducting properties, becoming a normal conductor.
Type I superconductors have a sharp transition from the superconducting state to the normal state.
Type II superconductors:
Type II superconductors have two critical magnetic fields: the lower critical field (Hc1) and the upper critical field (Hc2).
Below Hc1, the superconductor behaves as a perfect diamagnet, expelling magnetic field lines.
Between Hc1 and Hc2, known as the mixed state, the superconductor allows some magnetic field lines to penetrate in the form of quantized vortices.
Above Hc2, the superconductor loses its superconducting properties and becomes a normal conductor.
Type II superconductors have a more gradual transition from the superconducting state to the normal state.
Mechanism of Cooper pair formation:
Cooper pairs are the fundamental building blocks of superconductivity. They are formed by the interaction between electrons and lattice vibrations (phonons). The process can be explained as follows:
In a normal conductor, electrons experience scattering due to lattice imperfections, impurities, and thermal vibrations.
In a superconductor, at low temperatures, the lattice vibrations create a "glue" or attractive force between electrons.
When an electron moves through the lattice, it slightly distorts the lattice and creates a positive charge imbalance (a "hole") behind it.
Another electron is attracted to this positive charge imbalance and follows behind, creating a correlated motion.
The lattice vibrations (phonons) mediate this attractive interaction between the electrons, leading to the formation of Cooper pairs.
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5. (25 points) OPTIONAL PROBLEM. You are given one of the small mirrors that we used in the lab demonstrations, so it has both a convex side and a concave side. The magnitude of the radius of curvature is 18.0 cm for both sides. a. (10 points) You put an object that is 5.0 cm tall in front of the mirror's CONCAVE side. An image is formed 6.0 cm behind the mirror. Determine: i. (5 pts) The location of the object- i.e., the object distance. (2 pts) The size of the image. (1 pt) The type of the image: Real or Virtual. To get credit, you must briefly justify your choice. A "bare" answer will not get any credit. (1 pt) The orientation of the image: Upright or Inverted. To get credit, you must briefly justify your choice. A "bare" answer will not get any credit. (1 pt) The magnification of the image (give a value). ii. iii. iv. V.
Answer: (1) object distance = -18cms
(2)Size = 1.67cms.
(3)Image: real
(4)Orientation: upright
(5)magnification = 1/3
Magnitude of the radius of curvature = 18.0 cm
Object height, h = 5.0 cm
Image distance, v = -6.0 cm (negative because the image is formed on the same side of the object)
1) Object distance: 1/f = 1/v - 1/u
Where, f = focal length of the mirror. For a spherical mirror, the focal length is given by:
f = R/2 Where, R = radius of curvature of the mirror.
For a concave mirror, the focal length is negative. R = -18.0 cm, f = -9.0 cmv = -6.0 cm
1/-9 = 1/-6 - 1/u1/u
= 1/-9 + 1/-6u
= -18.0 cm (negative because the object is placed on the same side of the mirror as the image)
Therefore, the object distance is -18.0 cm.
2) Size of the image, h' = ?
The magnification of the mirror is given by:
m = -v/u Where, m = magnification of the image. For a concave mirror, the magnification is negative. v = -6.0 cm, u = -18.0 cm. m = -6/-18 = 1/3This means that the image is one-third the size of the object.
h' = m × hh' = (1/3) × 5.0h' = 1.67 cm.
Therefore, the size of the image is 1.67 cm.
3) Type of image: the image is formed on the same side of the mirror as the object. Therefore, the image is virtual.
4) Orientation of the image: The magnification is positive, which means that the image is upright.
5) Magnification of the image, m = ?We have already calculated the magnification of the image, which is:
m = -v/u = -(-6)/(-18) = 1/3.
Therefore, the magnification of the image is 1/3.
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An the light emitted from electronic transition in a H atom for a transition from n = 3 to n = 2 has a characteristic wavelength range of 656 nm. Calculate the following: The frequency of the light em
Therefore, the frequency of the light emitted from this electronic transition is approximately 4.58 × 10^14 s^-1 (or hertz).
Electronic transition in a hydrogen atom for a transition from n = 3 to n = 2 has a characteristic wavelength range of 656 nm. To calculate the frequency of the light emitted, we can use the following equation: c = λν,where c is the speed of light, λ is the wavelength, and ν is the frequency. We are given the wavelength, so we can solve for the frequency:ν = c/λ = (3.00 × 10^8 m/s)/(656 nm × 10^-9 m/nm) ≈ 4.58 × 10^14 s^-1. Therefore, the frequency of the light emitted from this electronic transition is approximately 4.58 × 10^14 s^-1 (or hertz).
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In a baseball game, a batter hits the 0.150−kg ball straight Part A back at the pitcher at 190 km/h. If the ball is traveling at 150 km/h just before it reaches the bat, what is the magnitude of the average force exerted by the bat on it if the collision lasts 6.0 ms ? Express your answer with the appropriate units.
The magnitude of the average force exerted by the bat on the ball is approximately 1,500 N.
To find the magnitude of the average force exerted by the bat on the ball, we can use the impulse-momentum principle. The impulse experienced by an object is equal to the change in momentum it undergoes. In this case, the change in momentum of the ball is given by:
Δp = m * Δv, where Δp is the change in momentum, m is the mass of the ball, and Δv is the change in velocity. The initial velocity of the ball is 150 km/h, and the final velocity is -190 km/h (since it is traveling back towards the pitcher). Converting these velocities to m/s, we have: Initial velocity: 150 km/h = 41.7 m/s. Final velocity: -190 km/h = -52.8 m/s.
The change in velocity, Δv, is then (-52.8 m/s) - (41.7 m/s) = -94.5 m/s. Substituting the values into the equation for impulse, we have: Impulse = m * Δv = (0.150 kg) * (-94.5 m/s) = -14.18 kg·m/s. The magnitude of the average force, F, can be calculated using the equation: F = Δp / Δt, where Δt is the time interval of the collision.
Substituting the values, we have: F = (-14.18 kg·m/s) / (6.0 ms) = -2,363 N.
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A school bus is traveling at a speed of 0.2 cm/s. A school child on the bus launches a paper airplane, flying at 0.02 cm/s relative to the bus in the forward direction of the bus's motion. What is the speed of the paper airplane as seen by school children on the sidewalk through the bus windows? 0.248c 0.238c 0.219c 0.229c
The speed of the paper airplane as seen by school children on the sidewalk through the bus windows is approximately 0.229 times the speed of light (c).
To determine the speed of the paper airplane as seen by school children on the sidewalk through the bus windows, we need to consider the concept of relative velocities.
The velocity of an object can be calculated by adding or subtracting the velocities relative to different reference frames. In this case, the paper airplane's velocity is given relative to the bus, which is moving at a speed of 0.2 cm/s.
When the velocities are in the same direction, we can find the relative velocity by subtracting the magnitudes. Therefore, the relative velocity of the paper airplane with respect to the sidewalk is given by:
Relative velocity = Velocity of paper airplane - Velocity of bus
Relative velocity = 0.02 cm/s - 0.2 cm/s
Relative velocity = -0.18 cm/s
Since the relative velocity is negative, it means the paper airplane appears to move in the opposite direction of the bus's motion when observed by school children on the sidewalk through the bus windows.
To convert the relative velocity to a fraction of the speed of light (c), we divide the magnitude of the relative velocity by the speed of light:
Speed of paper airplane / Speed of light = |Relative velocity| / Speed of light
Speed of paper airplane / c = 0.18 cm/s / (2.998 x 10^10 cm/s)
Speed of paper airplane / c ≈ 0.229
Therefore, the speed of the paper airplane as seen by school children on the sidewalk through the bus windows is approximately 0.229 times the speed of light (c).
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A capacitor is a device used to store electric charge which allows it to be used in digital memory. a) Explain how a capacitor functions. b) What is the impact of using a dielectric in a capacitor? c) Define what is meant by electric potential energy in terms of a positively charge particle in a uniform electric field. d) How does electric potential energy relate to the idea of voltage?
a) When a voltage is applied, electrons accumulate on one plate and positive charge accumulates on the other, creating an electric field between the plates. b) The use of a dielectric in a capacitor increases its capacitance by reducing the electric field between the plates. c) Electric potential energy refers to the energy possessed by a positively charged particle in a uniform electric field. d)The potential energy of a charged particle in an electric field is proportional to the voltage across the field. As the voltage increases, the electric potential energy of the system increases, and vice versa.
a) A capacitor consists of two conductive plates separated by a dielectric material. When a voltage is applied across the plates, one plate accumulates an excess of electrons, becoming negatively charged, while the other plate accumulates a deficit of electrons, becoming positively charged. This creates an electric field between the plates. The dielectric material between the plates serves to increase the capacitance of the capacitor by reducing the electric field and allowing for a greater charge storage capacity.
b) The use of a dielectric in a capacitor has a significant impact on its performance. The dielectric material, often an insulating substance such as ceramic or plastic, increases the capacitance of the capacitor. It does this by reducing the electric field between the plates, which in turn reduces the voltage required to maintain a certain charge. The dielectric material has a high dielectric constant, which indicates its ability to store electrical energy. By increasing the dielectric constant, the charge storage capacity of the capacitor increases, making it more efficient in storing and releasing electric charge.
c) Electric potential energy refers to the energy possessed by a positively charged particle in a uniform electric field. When a positive charge is placed in an electric field, it experiences a force in the direction opposite to the field. To move the charge against this force, work must be done. This work is stored as potential energy in the system. The electric potential energy is directly proportional to the magnitude of the charge, the electric field strength, and the distance the charge is moved against the field.
d) Electric potential energy is closely related to the concept of voltage. Voltage is a measure of the electric potential difference between two points in an electric field. It represents the amount of work done per unit charge in moving a charge between those two points. The potential energy of a charged particle in an electric field is directly proportional to the voltage across the field. As the voltage increases, the potential energy of the system also increases. Therefore, voltage can be seen as a measure of the electric potential energy difference between two points in an electric field.
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The velocity of a longitudinal ultrasound wave in a diamond sample was measured at 64800 Km/h via Ultrasonic Inspection.
i. Calculate the dynamic Elastic Modulus of this material when its density is 3.5 g/cm³ and Poisson's ratio is 0.18.
ii. You have been asked to perform an Ultrasound investigation of a diamond component having access to one side of it. Which UT method are you going to use and why
iii. Calculate the velocity of a Shear wave (m/s) in this diamond sample.
The dynamic elastic modulus of a diamond sample was calculated to be 1552 GPa . The appropriate ultrasonic testing method for a diamond component investigation is pulse-echo using a normal probe. The velocity of a shear wave in the diamond sample was calculated to be 25995 m/s.
i. The dynamic elastic modulus (E) of the diamond sample can be calculated using the following formula:
E = ρv^2(1 - 2ν)
Substituting the given values, we get:
E = 3.5 g/cm^3 * (64800 km/h * 1000 m/km / 3600 s/h)^2 * (1 - 2*0.18)
E = 1552 GPa
Therefore, the dynamic elastic modulus of the diamond sample is 1552 GPa.
ii. The appropriate ultrasonic testing (UT) method for this diamond component would be the pulse-echo technique. This method involves sending a short pulse of ultrasound into the material from one side and detecting the reflected signal from the other side. The time delay between the transmitted and received signals can be used to determine the presence of any defects or anomalies.
iii. The velocity of a shear wave (vs) in the diamond sample can be calculated using the following formula:
vs = v / √(3(1-2ν))
Substituting the given values, we get:
vs = (64800 km/h * 1000 m/km / 3600 s/h) / √(3(1-2*0.18))
vs = 25995 m/s
Therefore, the velocity of a shear wave in the diamond sample is 25995 m/s.
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A plano-concave lens for an underwater camera is shown below. It's diopter under water is - 8.33. The radius of curvature of its front surface is 8 cm. Assuming that the index of fraction of water is 1.33, what is the index of fraction of the substance of which this lens it is made?
a. 2.00
b. 1.81
c. 1.52
d. 1.67
The index of refraction of the substance of which the lens is made is 1.81, which corresponds to option b.
The diopter under water is given as -8.33, which is equal to the reciprocal of the focal length in meters. Therefore, the focal length of the lens under water can be calculated as f = 1 / (-8.33) = -0.12 m.
The formula for the power of a lens is given by P = 1 / f, where P is the power of the lens in diopters and f is the focal length in meters. Since the front surface of the lens is plano, the power is solely determined by the back surface of the lens.
Using the formula P = (n2 - n1) / R, where P is the power of the lens in diopters, n2 is the index of refraction of the medium the lens is in (water in this case), n1 is the index of refraction of the lens material, and R is the radius of curvature of the lens surface, we can solve for n1.
Substituting the given values, -8.33 = (1.33 - n1) / (-0.08) and solving for n1, we get n1 = 1.81.
Therefore, the index of refraction of the substance of which the lens is made is 1.81, which corresponds to option b.
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A 58 Ni ion of charge 1 proton and mass 9.62x10-26kg is accelerated trough a potential difference of 3kV and deflected in a magnetic field of 0.12T. The velocity vector is perpendicular to the direction of magnetic field. a. [5] Find the radius of curvature of the ion's orbit. b. [4] A proton, accelerated to the same velocity as the 58Ni, also enters the same magnetic field. Is the radius of curvature of proton is going to be bigger/smaller/the same? Justify your answer.
(a) The radius of curvature of a 58Ni ion's orbit can be found using the given information of its charge, mass, potential difference, and magnetic field strength.
(b) The radius of curvature for a proton accelerated to the same velocity and entering the same magnetic field will be smaller than that of the 58Ni ion due to the proton's smaller mass.
(a) The radius of curvature (r) of the 58Ni ion's orbit can be determined using the equation r = (mv) / (qB), By substituting the given values and solving the equation, the radius of curvature can be calculated.
(b) For the proton, since it has a smaller mass compared to the 58Ni ion, its radius of curvature will be smaller. This can be justified by considering the equation r = (mv) / (qB). Therefore, the radius of curvature for the proton will be smaller than that of the 58Ni ion.
In conclusion, the radius of curvature for the 58Ni ion's orbit can be calculated using the given information, and the radius of curvature for the proton will be smaller due to its smaller mass.
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Imagine yourself as a NASA scientist who is planning the mission pf a new space probe. Choose which outer plant the probe will visit. Write a paragraph that defends the choice ( SCIENCE GRADE 6)
What would be the acceleration of gravity in the surface of a world with three times Earty's mans and in time radi? A planet's gravitational acceleration is given by A planet's gravitational acceleration given by 9, m2
Therefore, the acceleration due to gravity on this planet is 29.4 m/s².
The acceleration due to gravity at the surface of a planet is given by its mass and radius. The gravitational acceleration of a planet is expressed as:$$\text{Gravitational acceleration}=\frac{GM}{R^2}$$Where,G = Universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²M = Mass of the planetR = Radius of the planetOn the surface of the earth, the acceleration due to gravity is given by:$$g=\frac{GM}{R^2}$$Where,G = Universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²M = Mass of the earthR = Radius of the earthTherefore, the gravitational acceleration of the earth is:$$g=\frac{6.67×10^{-11}×5.98×10^{24}}{(6.38×10^6)^2}=9.8m/s^2$$We are given that the mass of the other planet is thrice that of the earth. Therefore, the gravitational acceleration on that planet can be found using the same equation, but with the mass being three times that of the earth. The radius of the planet is not given, but we can assume that it is the same as the earth. Therefore, the gravitational acceleration of the planet is:$$g=\frac{6.67×10^{-11}×3×5.98×10^{24}}{(6.38×10^6)^2}=\frac{9×9.8}{3}=29.4m/s^2$$Therefore, the acceleration due to gravity on this planet is 29.4 m/s².
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