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3 years ago

Descibe the real-world examples of Newton's third lawthat were idenified in "Applications of Newton's Laws."

Physics

2 answers:

ehidna [41]3 years ago

5 0

Answer:

A horse pulls a cart, a person walks on the ground

Explanation:

ValentinkaMS [17]3 years ago

5 0

Answer:

Know it's late but I couldn't find a good answer when I looked for it so i did it myself.

My answer:

One of the real world examples mentioned in the article was that of a rocket blasting off into space. It was explained that it's an example of Newton's third law because the rocket’s engines push on those gases (action force), and the gases push back (reaction force). This reaction shows an excellent example of Newton's third law.

Sample Response:

When gases escape downward out the tail of a rocket, they push upward on the rocket. When layers of the Earth collide, they force the Earth upward, forming mountains.

Explanation:

hope this helps <3

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Answer:

15 N

Explanation:

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f=1.5*10

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3 years ago

Read 2 more answers

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and

nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

$T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}$

k = 1.4

$T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K$

Work done is given as;

$W = \frac{1}{2} *m*(v_i^2 - v_e^2)$

inlet velocity is negligible;

$v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41 \ m/s$

Therefore, the exit velocity is 629.41 m/s

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3 years ago

a body is projected at an angle of 30 degrees to the horizontal with a velocity of 15m/s, calculate the time it takes to reach t

andreev551 [17]

Explanation:

The vertical component the velocity of the projectile is 15 m/s x sin 30 = 7.5 m/s.

The body is accelarating downwards at 10 m/s^2.

This means that every second its upward velocity reduces by 10 m/s.

So if the body is travelling upwards at 7.5 m/s then how long does it take for the velocity to become 0?

(7.5 m/s) / (10 m/s^2) = 0.75 s

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3 years ago