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sveta [45]
3 years ago
11

Which scientific law describes the relationship between action and reaction force pairs? Newton's third law of motion Kepler's l

aw of planetary motion Law of gravitation Law of conservation of mass
Physics
1 answer:
Hitman42 [59]3 years ago
7 0

Newton's third law of motion

Explanation:

Newton's third law of motion states that:

<em>"When an object A exerts a force on an object B (action force), then object B exerts an equal and opposite force (reaction force) on object A"</em>

It is important to note that this law is always valid, even when it seems it is not.

Consider for example the gravitational force that the Earth exerts on your body (= your weight). We can say that this is the action force. It may seems that there is no reaction force in this case. However, this is not true: in fact, your body also exerts an equal and opposite force on the Earth, and this is the reaction force. The reason that explains why we don't notice any effect on Earth due to this force is that the mass of the Earth is much larger than your mass, therefore the acceleration produced on the Earth because of the force you apply is negligible.

It is also important to note that the action-reaction pair of forces always act on two different objects, so they never appear in the same free-body diagram.

Learn more about Newton's third law of motion:

brainly.com/question/11411375

#LearnwithBrainly

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Gennadij [26K]

Answer:

E- The star becomes a red giant (LATEST STAGE)

F- The surface of the star becomes brighter and cooler

C- Pressure from the star's hydrogen-burning shell causes the non burning envelope to expand

A- The shell of hydrogen surrounding the star's nonburning helium core ignites.

D- The star's non burning helium core starts to contract and heat up

B- Pressure in the star's core decreases (EARLIEST STAGE)

(A star moves away from the main sequence once its core runs out of hydrogen to fuse into helium. The energy once supplied by hydrogen burning reduces and the core starts to compress under the force of gravity. This contraction allows the core and surrounding layers to heat up. Finally, the hydrogen shell around the core becomes hot enough to ignite hydrogen burning.

6 0
3 years ago
Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m
Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

A=0.0198

From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

d=0.0023

Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

d=0.0120

Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

A=0.0209

From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

Hence, This is the required solution.

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Answer:

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Explanation:

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Working in SI units, Power = Watts, Work = Joules, Time = seconds.

Power = 2240J/2.8s = 800 Watts.

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Answer:

Explanation:

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