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kicyunya [14]
3 years ago
7

A student pushes on a crate with the force of 100 Newtons directed to the right. What force does the crate exert on the student?

Note: It has to be done with the Newton's 2nd Law of Motion.
Physics
1 answer:
ale4655 [162]3 years ago
6 0

Answer: The Force exerted on the student by crate will be 100 N.

Explanation: As per the given Question student is trying to push the crate in right direction with the force of 100N.

And we know that, Newton's third law states that every action has equal and opposite reaction.

So from the Newton's 3rd law of motion it is very clear the student must be experiencing the same amount of force which he is applying.

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Leto [7]

Answer:

0.24 kg used up

Explanation:

He has a mass of 67 kg

The gravitational constant is 9.81 m/s^2

The distance upward is 3500 m

W = m*g*h

W = 67 * 9.81 * 3500

Work = 2,300,445 Joules

Work = 2300 kj

work = 2.30 * 10^6 joules in scientific notation.

Part B

He needs 4 times this amount to climb the mountain because the body is only 25% efficient in converting energy.

4*2.30 * 10^6 = 9.20 * 10^6 Joules of energy are therefore required.

The total amount in a kg of fat = 3.8 * 10^7 joules

x kg of fat is needed to provide 9.20.*10^6 joules

1 kg / (3.8 * 10^7 J ) = x kg / (9.20 * 10^6 J)

9.20 * 10 ^6 * 1 = 3.8 * 10^7 *x

9.20 * 10 ^6 / 3.8 * 10^7 = x

x = 0.24 kg of fat are needed

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3 years ago
Astronomers infer that the universe is expanding because distant galaxies all appear to ________. Astronomers infer that the uni
valina [46]

Answer:

Stretch and moving away from us, with more distant ones moving faster

Explanation:

Astronomers observed that the entire universe is getting bigger because distant galaxies appears to be moving away from us

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Answer:

When a particle or a system of particles move in a system where no external force acts, then the total linear momentum of the particle system remains constant.

Explanation:

Given data:

Total mass of the skateboarder, m_{1} = 70 \ kg

Mass of the friend, m_{2} = 50 \ kg

Initial velocity of the skateboarder, u_{1} = 3 \ m/s

Initial velocity of the the friend, u_{2} = 0

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From the conservation law of linear momentum,

m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v

70 \times 3 + 50 \times 0 = (70 + 50)v

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