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blondinia [14]
3 years ago
11

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

Physics
2 answers:
butalik [34]3 years ago
5 0

the answer you are looking for is 100N

pychu [463]3 years ago
4 0

Newton's 3rd law is: The force of body A on body B, is equal to the force of body B on body A.

This means that the force of the Student on the Crate (100N), is equal to the force of the Crate on the Student.

Which means the Crate also exerts a force of <u>100N</u> on the student.

---------------------------

Answer:

100N

-----------------------------

<em>Note: Newton's 3rd law can also be written as: </em>

<em>Every force has an opposite and equal reaction.</em>

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v=18.97 m/s

Explanation:

the equation for the above question is v²=u²+2as

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Efficiency of electrons causes negative charge why?​
Paraphin [41]

Answer:

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Explanation:

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3 years ago
7. A 1.0 kg metal head of a geology hammer strikes a solid rock with a velocity of 5.0 m/s. Assuming all the energy is retained
marta [7]

The increase in temperature of the metal hammer is 0.028 ⁰C.

The given parameters:

  • <em>mass of the metal hammer, m = 1.0 kg</em>
  • <em>speed of the hammer, v = 5.0 m/s</em>
  • <em>specific heat capacity of iron, 450 J/kg⁰C</em>

The increase in temperature of the metal hammer is calculated as follows;

Q = K.E\\\\mc \Delta T = \frac{1}{2}  mv^2\\\\\Delta T = \frac{v^2}{2 c}

where;

<em>c is the </em><em>specific heat capacity</em><em> of the metal hammer</em>

<em />

Assuming the metal hammer is iron, c = 450 J/kg⁰C

\Delta T = \frac{5^2}{2 \times 450} \\\\\Delta T = 0.028 \ ^0C

Thus, the increase in temperature of the metal hammer is 0.028 ⁰C.

Learn more about heat capacity here: brainly.com/question/16559442

8 0
3 years ago
What is solar thermal energy used to do—heat homes or power homes?
cluponka [151]

Answer:

A

Explanation:

7 0
4 years ago
A basketball player standing up with the hoop launches the ball straight up with an initial velocity of v_o = 3.75 m/s from 2.5
denis23 [38]

Answer:

a) The maximum height the ball will achieve above the launch point is 0.2 m.

b) The minimum velocity with which the ball must be launched is 4.43 m/s or 0.174 in/ms.

Explanation:

a)

For the height reached, we use 3rd equation of motion:

2gh = Vf² - Vo²

Here,

Vo = 3.75 m/s

Vf =  0m/s, since ball stops at the highest point

g = -9.8 m/s² (negative sign for upward motion)

h = maximum height reached by ball

therefore, eqn becomes:

2(-9.8m/s²)(h) = (0 m/s)² - (3.75 m/s²)²

<u>h = 0.2 m</u>

b)

To find out the initial speed to reach the hoop at height of 3.5 m, we again use 3rd eqn. of motion with h= 3.5 m - 2.5m = 1 m (taking launch point as reference), and Vo as unknown:

2(-9.8m/s²)(1 m) = (0 m/s)² - (Vo)²

(Vo)² = 19.6 m²/s²

Vo = √19.6 m²/s²

<u>Vo = 4.43 m/s</u>

Vo = (4.43 m/s)(1 s/1000 ms)(39.37 in/1 m)

<u>Vo = 0.174 in/ms</u>

<u />

6 0
3 years ago
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