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sweet [91]
3 years ago
5

A rocket is continuously firing its engines as it accelerates away from Earth. For the first kilometer of its ascent, the mass o

f fuel ejected is small compared to the mass of the rocket. For this distance, which of the following indicates the changes, if any, in the kinetic energy of the rocket, the gravitational potential energy of the Earth-rocket system, and the mechanical energy of the Earth-rocket system? System Gravitational System Potential Energy Rocket Kinetic Energy Mechanical Energy
(A) Increasing IncreasingIncreasing
(B) Increasing Increasing Constant IncreasingDecreasing Decreasing Decreasing Increasing Constant
Physics
1 answer:
melisa1 [442]3 years ago
3 0

Answer:

A) Increasing,  Increasing,  Increasing

Explanation:

The greater the rate of fuel ejection higher will be the kinetic energy of the rocket. As rocket is fire upward its fuel rejection rate increases and hence it's kinetic energy increases.

Gravitational potential energy,

as the rocket moves further it's r from the Earth increases. Hence the Gravitational potential energy increases as its height from the Earth increases.

Therefore,  mechanical energy of the rocket must also increase as it is sum of kinetic and potential energy.

Hence Option A is correct.

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Suppose you first walk 12.0 m in a direction 20? west of north and then 20.0 m in a direction 40.0? south of west. how far are y
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\left | \overrightarrow{A} \right |=12m \\ \\ \left | \overrightarrow{B} \right |=20m \\ \\ \theta_{A}=20^{\circ} \\ \\ \theta_{B}=40^{\circ}

From geometry, we know that:

\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}

Then using vector decomposition into components:

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Therefore:

R_x=A_x+B_x=-4.10-15.32=-19.42m \\ \\ R_y=A_y+B_y=11.27-12.85=-1.58m

So if you want to find out <span>how far are you from your starting point you need to know the magnitude of the vector \overrightarrow{R}, that is:
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\left | \overrightarrow{R} \right |=&#10;\sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}

Finally, let's find the <span>compass direction of a line connecting your starting point to your final position. What we are looking for here is an angle that is shown in Figure 2 which is an angle defined with respect to the positive x-axis. Therefore:

</span>\theta_R=180^{\circ}+tan^{-1}(\frac{\left | R_y \right |}{\left | R_x \right |}) \\ \\ \theta_R=180^{\circ}+tan^{-1}(\frac{1.58}{19.42}) \\ \\ \theta_R=180^{\circ}+4.65^{\circ}=185.85^{\circ}


6 0
3 years ago
What is Gravitational force?​
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