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Hitman42 [59]
3 years ago
15

Objects with masses of 135 kg and a 435 kg are separated by 0.500 m. Find the net gravitational force exerted by these objects o

n a 38.0-kg object placed midway between them.
Physics
1 answer:
IrinaVladis [17]3 years ago
8 0

Answer:

The net gravitational force on the mass is 1.27\times 10^{-5}

Explanation:

We have by Newton's law of gravity the force of attraction between masses m_{1},m_{2}

F_{att}=G\frac{m_{1}m_{2}}{r^{2}}

Applying vales we get

Force of attraction between 135 kg mass and 38 kg mass is

F_{1}=6.67\times 10^{-11}\frac{135\times 38}{(0.25)^{2}}\\\\F_{1}=5.47\times 10^{-6}N

Force of attraction between 435 kg mass and 38 kg mass is

F_{2}=6.67\times 10^{-11}\frac{435\times 38}{(0.25)^{2}}\\\\F_{2}=1.76\times 10^{-5}N

Thus the net force on mass 38.0 kg is F_{2}-F_{1}=1.76\times 10^{-5}-5.47\times 10^{-6}\\\\F_{2}-F_{1}=1.27\times 10^{-5}

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A box slides down a frictionless incline, gaining speed. The work done by the normal force n is _______.
jeka57 [31]

The work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

<h3>What is normal force?</h3>

The force of contact is called the normal force. When the two surfaces are in contact with each other, then the normal force acts.

This force is applied by the solid bodies on each other in order to prevent the passing through each other.

A box slides down a frictionless incline, gaining speed. For this box, the value of work done by normal force has to be found out. Let's analyze the given condition.

  • The body is gaining the speed, which means there is a change in kinetic energy.
  • The change in kinetic energy is equal to the work done.
  • The friction force is the product of coefficient of the friction and normal force.
  • The friction force for the given case is zero. Thus, the normal force must be equal to the zero.

Thus, the work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

Learn more about the normal force here;

brainly.com/question/10941832

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1 year ago
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a fly wheel of mass 12kg and radius 0.5m with wrapped around it.on the end of the rope is 2kg load.intially it is at rest.calcul
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Answer:

19.6 N of torque. The 2kg load is being affected by acceleration due to gravity which is 9.8 m/s^s

Explanation:

2×9.8=19.6

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What does malleablity means
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Able to be hammered or pressed permanently out of shape without breaking or cracking.

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Which of the following prefixes represents the largest value
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C should be right ok



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3 years ago
a bullet with a mass of 4.0g and a speed of 650m/s is fired at a block of wood with a mass of 0.095kg. the block rests on a fric
n200080 [17]

Part a)

Here in this we can use momentum conservation as there is no external force on it

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} +m_2v_{2f}

here we know that

m_1 = 0.004 kg

v_{1i} = 650 m/s

m_2 = 0.095

v_{2i} = 0

v_{2f} = 23 m/s

now by above equation

0.004*650 + 0.095* 0 = 0.004*v + 0.095*23

2.6 + 0 = 0.004*v + 2.185

v = 103.75 m/s

Part b)

Final kinetic energy of the system

KE_f = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2

KE_f = \frac{1}{2}*0.004*(103.75^2) + \frac{1}{2}*(0.095)*23^2

KE_f = 46.65 J

Initial Kinetic energy of the system will be

KE_f = \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2

KE_f = \frac{1}{2}*0.004*(650^2) + \frac{1}{2}*(0.095)*0^2

KE_f = 845 J

So here kinetic energy is decreased for this system

final energy is less than initial energy

6 0
2 years ago
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