Answer: 1. .090km/h
2. I dont know
3. 1560
Explanation:
1. add all the numbers up and divide by 3 bc thats how many numbers there are.
2. I dont know
3. add all distances up
Answer:
The angular acceleration is ![\alpha = 0.4418 \ rad /s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%200.4418%20%5C%20rad%20%2Fs%5E2)
Explanation:
From the question we are told that
The angular speed is ![w_f = 45 \ rev / minutes = \frac{45 * 2 * \pi }{60 }= 4.713 \ rad/s](https://tex.z-dn.net/?f=w_f%20%20%3D%20%2045%20%5C%20%20rev%20%2F%20minutes%20%3D%20%20%5Cfrac%7B45%20%2A%20%202%20%2A%20%20%5Cpi%20%20%7D%7B60%20%7D%3D%204.713%20%5C%20%20rad%2Fs)
The angular displacement is ![\theta =4 \ rev = 4 * 2 * \pi = 25.14 \ rad](https://tex.z-dn.net/?f=%5Ctheta%20%3D4%20%20%5C%20rev%20%20%3D%20%204%20%2A%202%20%2A%20%20%5Cpi%20%3D%20%2025.14%20%5C%20rad)
From the first equation of motion we can define the movement of the record as
![w_f ^2 = w_o ^2 + 2 * \alpha * \theta](https://tex.z-dn.net/?f=w_f%20%5E2%20%20%3D%20%20w_o%20%5E2%20%20%2B%20%202%20%2A%20%20%5Calpha%20%2A%20%20%5Ctheta)
Given that the record started from rest ![w_o = 0](https://tex.z-dn.net/?f=w_o%20%20%3D%20%200)
So
![4.713^2 = 2 * \alpha * 25.14](https://tex.z-dn.net/?f=4.713%5E2%20%20%3D%20%202%20%2A%20%20%5Calpha%20%2A%20%2025.14)
![\alpha = 0.4418 \ rad /s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%200.4418%20%5C%20rad%20%2Fs%5E2)
Answer:
300m
Explanation:
step one
given data
initial speed u= 15m/s
final speed v= 45m/s
time taken to attain final speed= 10seconds
Step two:
Let us first solve for the acceleration
a= Δv/t
a= 45-15/10
a=30/10
a= 3m/s
applying the equation of motion
![v^2=u^2+2as](https://tex.z-dn.net/?f=v%5E2%3Du%5E2%2B2as)
substituting our given data
![45^2+15^2+2*3*s\\\\2025=225+6s\\\\](https://tex.z-dn.net/?f=45%5E2%2B15%5E2%2B2%2A3%2As%5C%5C%5C%5C2025%3D225%2B6s%5C%5C%5C%5C)
collect like terms
2025-225=6s
1800=6s
divide both sides by 6
s=1800/6
s=300m