Answer:
Option A
Explanation:
From the question we are told that:
Mass
Velocity
Generally the equation for momentum for Ball A is mathematically given by
Initial Momentum
Final Momentum
Therefore
Generally the equation for momentum for Ball B is mathematically given by
Initial Momentum
Final Momentum
Therefore
Option A
Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ = / vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m
Answer:
1020 km
Explanation:
A complete rotation of the wheel equals a distance of 1 circumference.
The circumference is
where <em>d</em> is the diameter of the wheel.
300,000 rotations =
In kilometers, this is = 1017876/1000 km = 1020 km
Angular frequency of pendulum is given by
for both pendulum we have
For other pendulum
now we have relate angular frequency given as
[tex\omega_1 - \omega_2 = 3.13 - 2.98 = 0.15 rad/s[/tex]
now time taken to become in phase again is given as
now number of oscillations complete in above time