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:
You would have to find the friction force of the rubber block which would be found with the equation of Normal force (mass*gravity) times cooeficient of friction which would give 8.82 N for the amount of friction and because you need more force than 8.82 N (assuming gravity is 9.8)
Answer:
4541.8 J
Explanation:
First we find the mass of benzene available
mass = density x volume
= 0.867 x 34.1
= 29.5647 g
Then we find the amount of heat transferred by two processes:
heat tranferred = heat lost during temp drop + heat lost during freezing
= mcΔT + mL
= 29.5647 x 1.74 x (20.8 - 5.5) + 29.5647 x 127
= 4541.7883434 J
= 4541.8 J
Answer:
Work to move on ion inside a cell = 1.168×10^-20J
Explanation:
Potential difference is a ratio of work done by a charge.
It is given as
◇V= Workdone ×charge= W/a
Workdone = V × q= (7.3×10^-2)×(1.6×10^-19)
W= 1.168×10^-20Joules
