Answer:
α = -0.01625 rad / s²
Explanation:
This is an exercise in angular kinematics, we can use the relation
w = w₀ + 2 α θ
linear and angular variables are related
v = w r
w = v / r
Let's reduce the magnitudes to the SI system
v₀ = 91 km / h (1000m / 1km) (1h / 3600s) = 25.278 m / s
v = 48 km / h = 13,333 m / s
θ = 75 rev (2π rad / 1 rev) = 471.24 rad
Let's find the angular velocities
w₀ = v₀ / r
w₀ = 25.278 / 0.78
w₀ = 32,408 rad / s
w = v / r
w = 13.333 / 0.78
w = 17.09 rad / s
we calculate the angular acceleration
α = (w- w₀) / 2θ
α = (17.09 - 32.408) / (2 471.24)
α = -0.01625 rad / s²
the negative sign indicates that the wheel is stopping
Answer: ask god in your prayers
Explanation: im here for the same reason you are
I DONT KNOW THE ANSWER
Answer:
23.0 s
Explanation:
Given:
v₀ = 0 m/s
v = 19.8 m/s
a = 4.80 m/s²
Find: Δx and t
v² = v₀² + 2aΔx
(19.8 m/s)² = (0 m/s)² + 2 (4.80 m/s²) Δx
Δx = 40.84 m
v = at + v₀
19.8 m/s = (4.80 m/s²) t + 0 m/s
t = 4.125 s
The elevator takes 40.84 m and 4.125 s to accelerate, and therefore also 40.84 m and 4.125 s to decelerate.
That leaves 291.3 m to travel at top speed. The time it takes is:
291.3 m / (19.8 m/s) = 14.71 s
The total time is 4.125 s + 14.71 s + 4.125 s = 23.0 s.
Answer: 72200
Explanation:
First you must find the height for this is on an inclined hill using:
h=Lsin(angle) —> 28.0sin(11.0) = 5.34
Now you would just use the PE equation (mgh) because you are finding ME and when you starting from the top KE=0, showing that what ever answer you get from PE would equal the same for ME.
Using mgh:
m=1380
g=9.80
h=5.34
(1380)(9.8)(5.34)
=72218.16
*Rounding to the 3rd=72200
Hope this helps :)
Answer:
The angular acceleration of the centrifuge is -231.74 rad/s².
Explanation:
Given that,
Angular speed = 3500 rev/min = 366 rad/s
We need to calculate the angular displacement
Using formula of angular displacement
![\theta=2\pi n](https://tex.z-dn.net/?f=%5Ctheta%3D2%5Cpi%20n)
Put the value into the formula
![\theta=2\pi\times46](https://tex.z-dn.net/?f=%5Ctheta%3D2%5Cpi%5Ctimes46)
![\theta=289.02\ rad](https://tex.z-dn.net/?f=%5Ctheta%3D289.02%5C%20rad)
We need to calculate the angular acceleration
Using equation of motion
![\omega_{f}^2=\omega_{i}^2+2\alpha\theta](https://tex.z-dn.net/?f=%5Comega_%7Bf%7D%5E2%3D%5Comega_%7Bi%7D%5E2%2B2%5Calpha%5Ctheta)
![\alpha=\dfrac{\omega_{f}^2-\omega_{i}^2}{2\theta}](https://tex.z-dn.net/?f=%5Calpha%3D%5Cdfrac%7B%5Comega_%7Bf%7D%5E2-%5Comega_%7Bi%7D%5E2%7D%7B2%5Ctheta%7D)
![\alpha=\dfrac{0-(366)^2}{2\times289.02}](https://tex.z-dn.net/?f=%5Calpha%3D%5Cdfrac%7B0-%28366%29%5E2%7D%7B2%5Ctimes289.02%7D)
![\alpha=-231.74\ rad/s^2](https://tex.z-dn.net/?f=%5Calpha%3D-231.74%5C%20rad%2Fs%5E2)
Hence, The angular acceleration of the centrifuge is -231.74 rad/s².