We are given
the torque requirement of 97 Newton meter.
The formula of the torque is
τ = r * F * sinθ
where
τ is the torque
r = radius from the axis of rotation to the point of application.
F = force exerted
θ = the angle between the lever arm and the radius
Try to substitute the given and solve for F.
Answer:
A) F = - 8.5 10² N, B) I = 21 N s
Explanation:
A) We can solve this problem using the relationship of momentum and momentum
I = Δp
in this case they indicate that the body rebounds, therefore the exit speed is the same in modulus, but with the opposite direction
v₀ = 8.50 m / s
v_f = -8.50 m / s
F t = m v_f -m v₀
F =
let's calculate
F =
F = - 8.5 10² N
B) let's start by calculating the speed with which the ball reaches the ground, let's use the kinematic relations
v² = v₀² - 2g (y- y₀)
as the ball falls its initial velocity is zero (vo = 0) and the height upon reaching the ground is y = 0
v =
calculate
v =
v = 14 m / s
to calculate the momentum we use
I = Δp
I = m v_f - mv₀
when it hits the ground its speed drops to zero
we substitute
I = 1.50 (0-14)
I = -21 N s
the negative sign is for the momentum that the ground on the ball, the momentum of the ball on the ground is
I = 21 N s
Answer:
Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.
Explanation:
Since Juan is closer to the center and Kuri is away from the center so we can say that Juan will move smaller distance in one complete revolution
As we know that the distance moved in one revolution is given as

also the time period of revolution for both will remain same as they move with the time period of carousel
Now we can say that the speed is given as

so Juan will have less tangential speed. so correct answer will be
Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.
There are other forces at work here nevertheless we will imagine
it is just a conservation of momentum exercise. Also the given mass of the
astronaut is light astronaut.
The solution for this problem is using the formula: m1V1=m2V2 but
we need to get V1:
V1= (m2/m1) V2
V1= (10/63) 12 = 1.9 m/s will be the final speed of the astronaut after
throwing the tank.