Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height
of 30.0 m. During the collision at the bottom of the elevator shaft, a 86.0 kg passenger is stopped in 5.00 ms. (Assume that neither the passenger nor the cab rebounds.) What are the magnitudes of the (a) impulse and (b) average force on the passenger during the collision
1 answer:
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Answer:
Power, P = 600 watts
Explanation:
It is given that,
Mass of sprinter, m = 54 kg
Speed, v = 10 m/s
Time taken, t = 3 s
We need to find the average power generated. The work done divided by time taken is called power generated by the sprinter i.e.
Work done is equal to the change in kinetic energy of the sprinter.
P = 900 watts
So, the average power generated by the sprinter is 900 watts. Hence, this is the required solution.
The correct answer is:
Let's see why.
1 amu corresponds to the mass of the proton, which is:
if we convert this into energy, using Einstein equivalence between mass and energy, we find:
Now we can convert it into electronvolts:
So, 1 amu = 934 MeV. Therefore, 3 amu corresponds to 3 times this value:
Let's see why.
1 amu corresponds to the mass of the proton, which is:
if we convert this into energy, using Einstein equivalence between mass and energy, we find:
Now we can convert it into electronvolts:
So, 1 amu = 934 MeV. Therefore, 3 amu corresponds to 3 times this value: