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Romashka [77]
3 years ago
5

An astronaut in space pushes a piece of equipment to get it into the correct position. What does Newton's third law of motion te

ll us about this? Question 5 options: The equipment would have accelerated much faster if the astronaut had applied a much greater force to it. The equipment will "push back" on the astronaut with the same amount of force. The equipment would have stayed in the same exact location indefinitely until the very moment the astronaut applied force to it. The equipment will continue moving in the same direction indefinitely unless another force is applied to stop it.
Physics
2 answers:
Agata [3.3K]3 years ago
7 0

Answer: C and D

The equipment would have stayed in the same exact location indefinitely until the very moment the astronaut applied force to it.

The equipment will continue moving in the same direction indefinitely unless another force is applied to stop it.

Explanation: According to Newton's first law of motion which state that; A body at rest will continue to be at rest, or in linear motion will continue to move in a straight line, unless an external force act on it.

The equipment would have stayed in the same exact location indefinitely until the very moment the astronaut applied force to it.

immediately the astronaut apply force to the object by pushing in, Newton's first law will be manifested in which the equipment will continue moving in the same direction indefinitely unless another force is applied to stop it.

Alisiya [41]3 years ago
7 0

Answer:

its the equipment will "push back" on the astronaut the the same amount of force. i just took the test and thats what it said was the right answer was

Explanation:

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Potential energy is the energy possessed by a body by virtue of its position or configuration . And energy is the capacity of doing work since power is work divided by time therefore change in potential energy divided by time is equal to the power of the object.
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Assuming all object are moving with the same velocity, witch statement describes the object with the largest intertia?
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Mass is the quantitative measure of inertia of any object.

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A tightrope walker more easily balances on a tightwire if his pole
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B) droops.

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At the top of a pole vault, an athlete actually can do work pushing on the pole before releasing it. Suppose the pushing force t
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The work done on the athlete is approximately 2.09 J

Explanation:

From the definition of the work done by a variable force:

\displaystyle{\int_{x_i}^{x_j}F(x)dx}

and substituting with the function of our problem:

\displaystyle{\int_{0}^{0.19}(140x-190x^2)dx\approx2.09\mathrm{J}}

5 0
4 years ago
Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s
frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀)  we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

    y₁ = v₀₁ t + ½ g t²

    y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

     t ’= (t-t₀)

    y₂ = v₀₂ t ’+ ½ g t’²

    y₂ = 0 + ½ g (t-t₀)²

    y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

    S = y₁ -y₂

    S = ½ g t²– ½ g (t-t₀)²

    S = ½ g [t² - (t²- 2 t to + to²)]  

    S = ½ g (2 t t₀ - t₀²)

    S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to.  The rock y has not left and the distance increases

For t> = to.  the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t   v₁> v₂

3 0
4 years ago
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