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zheka24 [161]
3 years ago
10

A worker assigned to the restoration of the Washington Monument is checking the condition of the stone at the very top of the mo

nument. A nickel with the mass of 0.005 kg is in her shirt pocket. What is the kinetic energy (KE) of the nickel in her shirt pocket at the top of the monument?
Physics
1 answer:
Mnenie [13.5K]3 years ago
7 0

Answer:

Your answer is: K.E = 8.3 J

Explanation:

If the height (h) = 169.2 meters (m) and the mass (m) is 0.005 kilograms (kg) the total energy will be kinetic energy which is equal to the potential energy.

K.E = P.E and also P.E equals to mgh

Then you substitute all the parameters into the formula  ↓

P.E = 0.005 × 9.81 × 169.2

P.E = 8.2908 J

So your answer is 8.2908 but if you round it is K.E = 8.3

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A 9.0-kg bowling ball on a horizontal, frictionless surface experiences a net force of 6.0 n. what will be its acceleration?
Vladimir [108]

This question involves the concepts of Newton's Second Law of Motion.

The acceleration of the bowling ball will be "0.67 m/s²".

<h3>Newton's Second Law of Motion</h3>

According to Newton's Second Law of Motion, when an unbalanced force is applied on an object, it produces an acceleration in it, in the direction of the applied force. This acceleration is directly proportional to the force applied and inversely proportional to the mass of the object. Mathematically,

F=ma\\\\a=\frac{F}{m}

where,

  • a = acceleration = ?
  • F = Magnitude of the applied force = 6 N
  • m = Mass of the ball = 9 kg

Therefore,

a=\frac{6\ N}{9\ kg}

a = 0.67 m/s²

Learn more about Newton's Second Law of Motion here:

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2 years ago
It is an employer's responsibility to provide training for special equipment. true false
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Answer:

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6 0
3 years ago
Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.
rjkz [21]

Answer:

Momentum of block B after collision =-50\ kg\ ms^{-1}

Explanation:

Given

Before collision:

Momentum of block A = p_{A1}= -100\ kg\ ms^{-1}

Momentum of block B = p_{B1}= -150\ kg\ ms^{-1}

After collision:

Momentum of block A = p_{A2}= -200\ kg\ ms^{-1}

Applying law of conservation of momentum to find momentum of block B after collision p_{B2}.

p_{A1}+p_{B1}=p_{A2}+p_{B2}

Plugging in the given values and simplifying.

-100-150=-200+p_{B2}

-250=-200+p_{B2}

Adding 200 to both sides.

200-250=-200+p_{B2}+200

-50=p_{B2}

∴ p_{B2}=-50\ kg\ ms^{-1}

Momentum of block B after collision =-50\ kg\ ms^{-1}

6 0
3 years ago
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