Answer:
a) F = 21.16 N, b) a = 3.17 10²⁸ m / s
Explanation:
a) The outside between the alpha particles is the electric force, given by Coulomb's law
F = 
in that case the two charges are of equal magnitude
q₁ = q₂ = 2q
let's calculate
F = 
F = 21.16 N
this force is repulsive because the charges are of the same sign
b) what is the initial acceleration
F = ma
a = F / m
a = 
21.16 / 4.0025 1.67 10-27
a = 3.17 10²⁸ m / s
this acceleration is in the direction of moving away the alpha particles
Answer:
its speed is insignificant before the diver's speed change, so the result does not change
Explanation:
In this exercise of conservation of the momentum, the system is formed by the diver and the Earth
initial instant (before jumping)
p₀ = 0
final instant (after jumping)
= m v + M v²
how momentum is conserved
p₀ = p_{f}
0 = m v + M v²
v² = m / M v
since the mass of the Earth is M = 10²⁴ kg
its speed is insignificant before the diver's speed change, so the result does not change
The toy rocket is launched vertically from ground level, at time t = 0.00 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 72 m and acquired a velocity of 30 m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground with negligible air resistance.
The total energy of the rocket, which is a sum of its kinetic energy and potential energy, is constant.
At a height of 72 m with the rocket moving at 30 m/s, the total energy is m*9.8*72 + (1/2)*m*30^2 where m is the mass of the rocket.
At ground level, the total energy is 0*m*9.8 + (1/2)*m*v^2.
Equating the two gives: m*9.8*72 + (1/2)*m*30^2 = 0*m*9.8 + (1/2)*m*v^2
=> 9.8*72 + (1/2)*30^2 = (1/2)*v^2
=> v^2 = 11556/5
=> v = 48.07
<span>The velocity of the rocket when it impacts the ground is 48.07 m/s</span>
