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nata0808 [166]
3 years ago
7

There is a decrease in power. It can be concluded that there has also been a change in: A)current or voltage.

Physics
1 answer:
liraira [26]3 years ago
5 0

A current or voltage

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Pushing molecules like a wave.

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An electron is placed at a position on the x-axis where the electric potential is + 10 V. Which idea below best describes the fu
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option (e)

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As the electron is placed at a point where the potential is  + 10 V. electrons moves from lower potential to higher potential, so we cannot predict the direction of motion because we don't have any information about the potential at other point.

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During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be lau
Thepotemich [5.8K]

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T=6.75s

Explanation:

We must separate the motion into two parts, the first when the rocket's engines is on  and the second when the rocket's engines is off. So, we need to know the height (h_1) that the rocket reaches while its engine is on and we need to know the distance (h_2) that it travels while its engine is off.

For solving this we use the kinematic equations:

In the first part we have:

h_1=v_0T+\frac{1}{2}aT^2\\h_1=0*T+\frac{1}{2}(16\frac{m}{s^2})T^2\\h_1=8\frac{m}{s^2}T^2\\

and the final speed is:

v_f=v_0+aT\\v_f=0+16\frac{m}{s^2}T\\v_f=16\frac{m}{s^2}T

In the second part, the final speed of the first part it will be the initial speed, and the final speed is zero, since gravity slows it down the rocket.

So, we have:

v_f^2=v_0^2+2gh_2\\2gh_2=v_f^2-v_0^2\\h_2=\frac{v_f^2-v_0^2}{2g}\\h_2=\frac{0^2-(16\frac{m}{s^2}T)^2}{2(-9.8\frac{m}{s^2})}\\h_2=\frac{-256\frac{m^2}{s^4}T^2}{-19.6\frac{m}{s^2}}\\h_2=13.06\frac{m}{s^2}T^2

The sum of these heights will give us the total height, which is known:

h=h_1+h_2\\960m=8\frac{m}{s^2}T^2+13.06\frac{m}{s^2}T^2\\960m=21.06\frac{m}{s^2}T^2\\T^2=\frac{960m}{21.06\frac{m}{s^2}}\\T^2=45.58s^2\\T=\sqrt{45.58s^2}\\T=6.75s

This is the time that its needed in order for the rocket to reach the required altitude.

5 0
3 years ago
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