Answer:
α = (ω²)/8π
Explanation:
The angular acceleration(α) of the carousel can be determined by using rotational
kinematics:
ω² =ωo² + 2αθ
Let's make α the subject of this equation ;
ω² - ωo² = 2αθ
α = (ω² −ωo²)/2θ
Now, from the question, since initially at rest, thus, ωo = 0
Also,since 2 revolutions, thus, θ = 2 x 2π = 4π since one revolution is 2π
Plugging in the relevant values to get ;
α = (ω²)/2(4π)
α = (ω²)/8π
Speed is not a vector so horizontal speed does not sighify anything. If u meant velocity it vill be same as the initial velocity in horizontal direction
The answer for this change in the magnitude of momentum is the same for both because momentum is always conserved so both vehicles have the identical change.
So for determining who has the greater change in kinetic energy, momentum (P) = mv so P^2 = m^2 v^2 P^2 / 2m = 1/2 m v^2 = energy So the weightier the mass the smaller the energy change for the same momentum change so in here, the car has a greater change in kinetic energy.
Answer:
The speed in the first point is: 4.98m/s
The acceleration is: 1.67m/s^2
The prior distance from the first point is: 7.42m
Explanation:
For part a and b:
We have a system with two equations and two variables.
We have these data:
X = distance = 60m
t = time = 6.0s
Sf = Final speed = 15m/s
And We need to find:
So = Inicial speed
a = aceleration
We are going to use these equation:


We are going to put our data:


With these equation, you can decide a method for solve. In this case, We are going to use an egualiazation method.



![[\sqrt{(15m/s)^2-(2*a*60m)}]^{2}=[15m/s-(a*6s)]^{2}](https://tex.z-dn.net/?f=%5B%5Csqrt%7B%2815m%2Fs%29%5E2-%282%2Aa%2A60m%29%7D%5D%5E%7B2%7D%3D%5B15m%2Fs-%28a%2A6s%29%5D%5E%7B2%7D)








If we analyze the situation, we need to have an aceleretarion greater than cero. We are going to choose a = 1.67m/s^2
After, we are going to determine the speed in the first point:




For part c:
We are going to use:




Given:
Distance = 50 yard = 45.72 meter
Speed = 40 km/hr = 11.11 m/s
To find:
Time required by ball to reach the receiver = ?
Formula used:
speed = 
Solution:
The speed of the ball is given by,
speed = 
Thus,
Time = 
Distance = 50 yard = 45.72 meter
Speed = 40 km/hr = 11.11 m/s
Time = 4.12 second
Hence, ball reaches the receiver in 4.12 second.