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:
(a) T = 0.015 N
(b) M = 1.53 x 10⁻³ kg = 1.53 g
Explanation:
(a) T = 0.015 N
First, we will find the speed of waves:
where,
v = speed of wave = ?
f = frequency = 120 Hz
λ = wavelength = 6 cm = 0.06 m
Therefore,
v = (120 Hz)(0.06 m)
v = 7.2 m/s
Now, we will find the linear mass density of the coil:
where,
μ = linear mass density = ?
m = mass = 1.45 g = 1.45 x 10⁻³ kg
l = length = 5 m
Thereforre,
Now, for the tension we use the formula:
<u>T = 0.015 N</u>
<u></u>
(b)
The mass to be hung is:
<u>M = 1.53 x 10⁻³ kg = 1.53 g</u>