Answer:
Maximum contact force = 487.4 N
Explanation:
The graph showing force vs time is attached to the solution of this question.
According to the an interpretation of Newton's second law of motion, the magmitude of the change in momentum is equal to the magnitude of the impulse experienced duting this impact.
F.t = Δ(momemtum)
Δ(momemtum) = 0.044 [36 - (-36)] = 0.044 × 72 = 3.168 kgm/s
Meaning, Impulse = 3.168 kgm/s too.
The area under the curve above is the impulse we just obtained.
If the maximum contact force is called F
The impulse = Area under this graph
= (1/2)(F)(4-0) + (F)(7-4) + (1/2)(F)(10-7)
= 2F + 3F + 1.5F = (6.5F) mN.s = (0.0065F) N.s
0.0065F = 3.168
F = (3.168/0.0065) = 487.4 N
Hope this Helps!!!
To solve this problem we will apply the concepts related to Coulomb's law for which the Electrostatic Force is defined as,
Here,
k = Coulomb's constant
= Charge at each object
r = Distance between them
As the distance is doubled so,
Therefore the factor is 1/4
Answer:
D. 5.7 m
Explanation:
The acceleration of the car is -8.7 m/s² and the car has an initial speed (u) of 10 m/s. The car said to come to rest, that means that the final velocity (v) of the car is 0 m/s.
To find the distance traveled by the car (s) before complete stop, this equation is being used:
v² = u² + 2as
The car traveled 5.7 m before coming to a complete stop