The horizontal force applied is 160 N while the velocity is 2.03 m/s.
<h3>What is the speed of the car?</h3>
The work done by the car is obtained as the product of the force and the distance;
W = F x
F = ?
x = 30.0 m
W = 4,800 J
F = 4,800 J/30.0 m
F = 160 N
But F = ma
a = F/m
a = 160 N/2.30 ✕ 10^3-kg
a= 0.069 m/s
Now;
v^2 = u^2 + 2as
u = 0/ms because the car started from rest
v = √2as
v = √2 * 0.069 * 30
v = 2.03 m/s
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Answer:
You need to use charts or a calculator to get 56%
Explanation:
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Answer:
The value of the spring constant of this spring is 1000 N/m
Explanation:
Given;
equilibrium length of the spring, L = 10.0 cm
new length of the spring, L₀ = 14 cm
applied force on the spring, F = 40 N
extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm
From Hook's law
Force applied to a spring is directly proportional to the extension produced, provided the elastic limit is not exceeded.
F ∝ e
F = ke
where;
k is the spring constant
k = F / e
k = 40 / 0.04
k = 1000 N/m
Therefore, the value of the spring constant of this spring is 1000 N/m
