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Art [367]
4 years ago
7

which object has the most gravitational potential energy? A. an 8 kg book at a height of 3m B. an 5 kg book at a height of 3 m C

. A 5 kg book at a height of 2 m D. a book at a height of 2 m
Physics
2 answers:
astra-53 [7]4 years ago
6 0
I think A is the correct answer because its high is more higher compared to the others, and the mass really does not matter, to know the gravitational potential energy, we need to know how high the object is located because gravity does not show any favor to an object that has more mass or an object that doesnt
amid [387]4 years ago
5 0

Answer: an 8 kg book at a height of 3m

Explanation:

Potential energy is the energy possessed by an object by virtue of its position.

P.E=m\times g\times h

m= mass of object = 200 kg

g = acceleration due to gravity = 10ms^{-2}

h = height of an object  

1. an 8 kg book at a height of 3m

P.E=8kg\times 10ms^{-2}\times 3m=240Joules

2. an 5 kg book at a height of 3 m

P.E=5kg\times 10ms^{-2}\times 3m=150Joules

3. an 5 kg book at a height of 2 m

P.E=5kg\times 10ms^{-2}\times 2m=100Joules

4. a book at a height of 2 m

P.E=0kg\times 10ms^{-2}\times 3m=0Joules

Thus a 8 kg book at a height of 3 m has most potential energy.

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An undamped oscillator has period ro= 1.000 s, but I now add a little damping so that its period changes to r i= 1.001 s.
Firlakuza [10]

This Question has mistakes.Correct question is

An undamped oscillator has period  τo=1.000s , but I now add a little damping so that its period changes to  τ1=1.001s.

What is the damping factor β? By what factor will the amplitude of oscillation decrease after 10 cycles? Which effect of damping would be more noticeable, the change of period or the decrease of the amplitude?

Answer:

β=0.28 /sec

Amplitude=0.0606

Explanation:

T_{o}=2\pi  /w\\T_{1}=2\pi /\sqrt{w^{2} -\beta^{2}  }\\ As\\T_{o}=1 \\T_{1}=1.001\\From  T_{o}\\w=2\pi \\So\\T_{1}=w/\sqrt{w^{2} -\beta^{2}  }\\\beta =0.00447w\\\beta =0.00447(2\pi )\\\beta =0.28sec^{-1}\\ Amplitude=Ae^{-\beta t}\\ after\\ t=10T_{1}\\so\\Ae^{-\beta(10t_{1}) }=e^{-\beta(10t_{1}) }\\=e^{-\(0.28)(10)(1.001) }\\Amplitude=0.0606

7 0
4 years ago
Why would researchers not be allowed to recreate the Little Albert experiment today?
wariber [46]

Answer:

Explanation:

En la historia de la ciencia se han dado auténticas barbaridades. Pruebas con animales que hoy no perdonaría nadie, o investigaciones de conducta con personas como la de la cárcel de Stanford, que se han saldado como una especie de pasado incómodo sobre los límites de la experimentación. Sin embargo, pocos se pueden acercar por su carácter perturbador al denominado experimento de Little Albert o Pequeño Albert: El salvaje intento por probar con un bebé que las fobias pueden ser condicionadas y aprendidas. Y lo que es peor, conseguirlo.

Esta idea surgió de la mente de John Broadus Watson, reconocido padre de la rama conductista de la psicología, que desde 1913 había comenzado a probar en animales sus tesis. Estas bebían directamente del los estudios de Iván Pavlov, fisiólogo ruso que ganó el Nobel en 1904 por sus estudios sobre el sistema digestivo, pero que también sentó precedentes sobre la psicología.

link por si te interesa:

https://hipertextual.com/2017/10/pequeno-albert

3 0
3 years ago
A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for
nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block (W), measured in newtons, is:

W = m\cdot g (1)

Where:

m - Mass of the block of ice, measured in kilograms.

g  - Gravitational acceleration, measured in meters per square second.

If we know that m = 14\,kg and g = 9.807\,\frac{m}{s^{2}}, the magnitudes of the weight and normal force of the block of ice are, respectively:

N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

N = W = 137.298\,N

And the pull force is:

F_{pull} = 98\,N

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f} (2)

Where:

v_{o}, v_{f} - Initial and final speeds of the block, measured in meters per second.

\Sigma F - Horizontal net force, measured in newtons.

\Delta t - Impact time, measured in seconds.

Now we clear the final speed in (2):

v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}

If we know that v_{o} = 0\,\frac{m}{s}, m = 14\,kg, \Sigma F = 98\,N and \Delta t = 1.40\,s, then final speed of the ice block is:

v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}

v_{f} = 9.8\,\frac{m}{s}

The final speed of the block of ice is 9.8 meters per second.

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