Problem: Hooke's law states that the force on a spring varies directly with the distance that it is stretched. If a spring has a
1 answer:
Answer:
Restoring force of the spring is 50 N.
Explanation:
Given that,
Spring constant of the spring, k = 100 N/m
Stretching in the spring, x = 0.5 m
We need to find the restoring force of the spring. It can be calculated using Hooke's law as "the force on a spring varies directly with the distance that it is stretched".
F = 50 N
So, the restoring force of the spring is 50 N. Hence, this is the required solution.
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The force between the two objects is 19.73 nN.
<u>Explanation: </u>
Any force acting between two objects tends to be directly proportional to the product of their masses and inversely proportional to the square of the distance between the two objects. And this kind of attraction force between two objects is termed as gravitational force.
So if we consider and
as the masses of both objects and let d be the distance of separation of two objects. Then the force between the two objects can be determined as below:
As gravitational constant ,
= 20 kg and
= 100 kg, while d = 2.6 m, then
Thus, we get finally,
As we know, nano denoted by letter 'n' equals to
So the force acting between two objects is 19.73 nN.
Answer:
a) a = 3.72 m / s², b) a = -18.75 m / s²
Explanation:
a) Let's use kinematics to find the acceleration before the collision
v = v₀ + at
as part of rest the v₀ = 0
a = v / t
Let's reduce the magnitudes to the SI system
v = 115 km / h (1000 m / 1km) (1h / 3600s)
v = 31.94 m / s
v₂ = 60 km / h = 16.66 m / s
l
et's calculate
a = 31.94 / 8.58
a = 3.72 m / s²
b) For the operational average during the collision let's use the relationship between momentum and momentum
I = Δp
F Δt = m v_f - m v₀
F =
F = m [16.66 - 31.94] / 0.815
F = m (-18.75)
Having the force let's use Newton's second law
F = m a
-18.75 m = m a
a = -18.75 m / s²