Answer:
W = 55.12 J
Explanation:
Given,
Natural length = 6 in
Force = 4 lb, stretched length = 8.4 in
We know,
F = k x
k is spring constant
4 = k (8.4-6)
k = 1.67 lb/in
Work done to stretch the spring to 10.1 in.
W = 55.12 J
Work done in stretching spring from 6 in to 10.1 in is equal to 55.12 J.
Answer:
The velocity of the truck after this elastic collision is 15.7 m/s
Explanation:
It is given that,
Mass of the car,
Mass of the truck,
Initial velocity of the car,
Initial velocity of the truck, u₂ = 0
After the collision the velocity of the car is, v₁ = -11 m/s
Let v₂ is the velocity of the truck after this elastic collision. Using the conservation of momentum as :
So, the velocity of the truck after this elastic collision is 15.7 m/s. Hence, the correct option is (c).
Answer:
Explanation:
T = 2π
(T / 2π)² = L/g
g = 4π²L/T²
g = 4π²(0.75000)/(1.7357)²
g = 9.82814766...
g = 9.8281 m/s²