F = 52000 N
m = 1060 kg
a= F/m = 52000 N/1060 kg = 49.0566 m/s^2
Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m
Explanation:
Given that;
mass of vehicle m = 1000 kg
for a low speed test; V = 2.5 m/s
bumper maximum deflection = 4 cm = 0.04 m
First we determine the energy of the vehicle just prior to impact;
W_v = 1/2mv²
we substitute
W_v = 1/2 × 1000 × (2.5)²
W_v = 3125 J
now, the the effective design stiffness k will be:
at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;
hence;
W_v = 1/2kx²
we substitute
3125 = 1/2 × k (0.04)²
3125 = 0.0008k
k = 3125 / 0.0008
k = 3906250 N/m
Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m
Answer:à
Explanation:waves carry energy in the direction in which they move
Answer:
(7.8) x (9.8 m/s) = 76.44 m/s
during the time he spent falling.
Since his falling speed was zero when he 'stepped' off of the top,
he hit the ground at 76.44 m/s.
That's about 170 miles per hour.
I'll bet he left one serious crater!
I hope this helps too! :D
Explanation:
Answer:
6.13428 rev/s
Explanation:
= Final moment of inertia = 4.2 kgm²
I = Moment of inertia with fists close to chest = 5.7 kgm²
= Initial angular speed = 3 rev/s
= Final angular speed
r = Radius = 76 cm
m = Mass = 2.5 kg
Moment of inertia of the skater is given by

In this system the angular momentum is conserved

The rotational speed will be 6.13428 rev/s