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:
You might be interested in
Answer:
a) v = 1.075*10^7 m/s
b) FB = 7.57*10^-12 N
c) r = 10.1 cm
Explanation:
(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:
(1)
q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C
V: potential difference = 1.2*10^6 V
You replace the values of the parameters in the equation (1):
The kinetic energy of the particle is also:
(2)
m: mass of the particle = 6.64*10^⁻27 kg
You solve the last equation for v:
the sped of the alpha particle is 1.075*10^6 m/s
b) The magnetic force on the particle is given by:
B: magnitude of the magnetic field = 2.2 T
The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1
the force exerted by the magnetic field on the particle is 7.57*10^-12 N
c) The particle describes a circumference with a radius given by:
the radius of the trajectory of the electron is 10.1 cm