Answer:
T = 3.23 s
Explanation:
In the simple harmonic movement of a spring with a mass the angular velocity is given by
w = √ K / m
With the initial data let's look for the ratio k / m
The angular velocity is related to the frequency and period
w = 2π f = 2π / T
2π / T = √ k / m
k₀ / m₀ = (2π / T)²
k₀ / m₀ = (2π / 3.0)²
k₀ / m₀ = 4.3865
The period on the new planet is
2π / T = √ k / m
T = 2π √ m / k
In this case the amounts are
m = 6 m₀
k = 10 k₀
We replace
T = 2π√6m₀ / 10k₀
T = 2π √6/10 √m₀ / k₀
T = 2π √ 0.6 √1 / 4.3865
T = 3.23 s
<span>1/3
The key thing to remember about an elastic collision is that it preserves both momentum and kinetic energy. For this problem I will assume the more massive particle has a mass of 1 and that the initial velocities are 1 and -1. The ratio of the masses will be represented by the less massive particle and will have the value "r"
The equation for kinetic energy is
E = 1/2MV^2.
So the energy for the system prior to collision is
0.5r(-1)^2 + 0.5(1)^2 = 0.5r + 0.5
The energy after the collision is
0.5rv^2
Setting the two equations equal to each other
0.5r + 0.5 = 0.5rv^2
r + 1 = rv^2
(r + 1)/r = v^2
sqrt((r + 1)/r) = v
The momentum prior to collision is
-1r + 1
Momentum after collision is
rv
Setting the equations equal to each other
rv = -1r + 1
rv +1r = 1
r(v+1) = 1
Now we have 2 equations with 2 unknowns.
sqrt((r + 1)/r) = v
r(v+1) = 1
Substitute the value v in the 2nd equation with sqrt((r+1)/r) and solve for r.
r(sqrt((r + 1)/r)+1) = 1
r*sqrt((r + 1)/r) + r = 1
r*sqrt(1+1/r) + r = 1
r*sqrt(1+1/r) = 1 - r
r^2*(1+1/r) = 1 - 2r + r^2
r^2 + r = 1 - 2r + r^2
r = 1 - 2r
3r = 1
r = 1/3
So the less massive particle is 1/3 the mass of the more massive particle.</span>
No they don't. Incident rays parallel to the axis of a concave mirror
reflect from the mirror's surface and converge at its focal point.
Answer:
Explanation:
This problem can be solved with the conservation of the momentum.
If the ball is fired upward, the momentum before and after the ball is fired must conserve. Hence, the speed of the ball is the same that the speed of the car just in the moment in wich the ball is fired.
Hence, the result depends of the acceleration of the car. If the change in the speed is higher than the speed of the ball, it is probably that the ball will be behind the car or it will come back to the car.
If the ball is fired forward, and if the change in the speed of the car is not enogh, the ball will be in front of the car.
HOPE THIS HELPS!!
