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Aleksandr [31]
3 years ago
6

How does your power output in climbing the stairs compare to the power output of a 100-watt light bulb? if your power could have

been harnessed and the energy converted to electricity, how many 100-watt bulbs could you have kept burning during your climb?
Physics
2 answers:
boyakko [2]3 years ago
8 0

Answer:

since the estimate we made is very rough, we can say that the power output of the person is comparable to the power output of the light bulb of 100 W.

Explanation:

cricket20 [7]3 years ago
6 0
1) Assuming an adult person has an average mass of m=80 kg, and assuming it takes about 30 seconds to climb 5 meters of stairs, the energy used by the person is
E=mgh=(80 kg)(9.81 m/s^2)(5 m)=3924 J
So the power output is 
P= \frac{E}{t}= \frac{3924 J}{30 s} \sim 130 W

And since the estimate we made is very rough, we can say that the power output of the person is comparable to the power output of the light bulb of 100 W.

2) Based on the results we found in the previous part of the exercise, since the power output of the person is comparable to the power output of 1 light bulb of 100 W, we can say that the person could have kept burning only one 100-W light bulb during the climb.
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Answer:

$25

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As we know that

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Answer:

Let's investigate the case where the cable breaks.

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\vec{L}_1 = \vec{L}_2\\\vec{L}_1 = m\vec{v_0} \\\vec{L}_2 = I\vec{\omega}\\

The projectile embeds itself to the ball, so they can be treated as a combined object. <u>The moment of inertia of the combined object is equal to the sum of the moment of inertia of both objects. </u>

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v = \omega r

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F = \frac{m(\omega r)^2}{r} = m\omega^2 r = 300\\mv_0r  =I\omega\\\\\omega = mv_0r/I\\300 = m(\frac{mv_0r}{I})^2r = m(\frac{mv_0r}{2mr^2})^2r = m(\frac{v_0}{2r})^2r = \frac{mv_0^2r}{4r^2} = \frac{mv_0^2}{4r}\\300 = \frac{0.8v_0^2}{4r}\\1500 = v_0^2/r\\v_0 = \sqrt{1500r}

As can be seen, the maximum velocity for the projectile without breaking the cable is \sqrt{1500r}, where r is the length of the cable.

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3 years ago
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