"The distance between the front and rear wheels of a 2280 kg car is 3.15 m. The car’s center of gravity lies in the middle of th
e car, 1.05 m behind the front wheels. When the car is at rest, what is the ratio of the force exerted by the ground on the front wheels to that exerted on the rear wheels?"
1 answer:
Answer:
2
Explanation:
Hello to solve this problem we must initially draw the free body diagram of the car, and remember that for a body to be in balance the sum of its moments must be zero.
When this equation is raised, we can find the relationship between the forces of the front and rear wheel, and the complete procedure is attached.
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= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)
= -0.00225 m
New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248
D
Answer:
m = 1 kg
Explanation:
Given that,
The force constant of the spring, k = 39.5 N/m
The frequency of oscillation, f = 1 Hz
The frequency of oscillation is given by the formula as formula as follows :
So, the mass that is attached to the spring is 1 kg.