A. Two point charges totaling 7.50 µC exert a repulsive force of 0.100 N on one another when separated by 0.911 m. What is the c
1 answer:
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Answer:
The maximum height reached by the two blocks is approximately 0.1147959 × v₀²
Explanation:
The mass of block B = m
The mass of block A = 3·m
The initial velocity of block B, v₂ = 0 m/s
The initial velocity of block A, v₁ = v₀
The amount of friction between the blocks and the surface = Negligible friction
By the Law of conservation of linear momentum, we have;
Total initial momentum = Total final momentum
3·m·v₁ + m·v₂ = (3·m + m)·v₃ = 4·m·v₃
Plugging in the values for the velocities gives;
3·m × v₀ + m × 0 = (3·m + m)·v₃ = 4·m·v₃
∴ 3·m × v₀ = 4·m·v₃
The kinetic energy, K.E. of the combined blocks after the collision is given as follows;
K.E. = 1/2 × mass × v²
The potential energy, P.E., gained by the two blocks at maximum height = The kinetic energy, K.E., of the two blocks before moving vertically upwards
The potential energy, P.E. = m·g·h
Where;
m = The mass of the object at the given height
g = The acceleration due to gravity
h = The height at which the object of mass, 'm', is located
Therefore, for h = The maximum height reached by the two blocks, we have;
P.E. = K.E.
The maximum height reached by the two blocks, h ≈ 0.1147959·v₀².
Answer:
If the acceleration is constant, the movements equations are:
a(t) = A.
for the velocity we can integrate over time:
v(t) = A*t + v0
where v0 is a constant of integration (the initial velocity), for the distance traveled between t = 0 units and t = 10 units, we can solve the integral:
Where to obtain the actual distance you can replace the constant acceleration A and the initial velocity v0.