I think that the answer is c. good luck
Answer:
ωf = 13 rad/s
Explanation:
- The angular acceleration, by definition, is just the rate of change of the angular velocity with respect to time, as follows:
- α = Δω/Δt = (ωf-ω₀) / (tfi-t₀)
- Choosing t₀ = 0, and rearranging terms, we have
![\omega_{f} = \omega_{o} + \alpha *t (1)](https://tex.z-dn.net/?f=%5Comega_%7Bf%7D%20%3D%20%5Comega_%7Bo%7D%20%2B%20%5Calpha%20%2At%20%20%281%29)
where ω₀ = 5 rad/s, t = 4 s, α = 2 rad/s2
- Replacing these values in (1) and solving for ωf, we get:
![\omega_{f} = 5 rad/s + (2 rad/s2*4 s) = 13 rad/s (2)](https://tex.z-dn.net/?f=%5Comega_%7Bf%7D%20%3D%205%20rad%2Fs%20%2B%20%282%20rad%2Fs2%2A4%20s%29%20%3D%2013%20rad%2Fs%20%282%29)
- The wheel's angular velocity after 4s is 13 rad/s.
The electrons in a piece of metal will leave when the polarized magnesis effect is reppeled and heated up to a certain tempeture.
![\star](https://tex.z-dn.net/?f=%5Cstar)
![\blue\star](https://tex.z-dn.net/?f=%5Cblue%5Cstar)
<h3>
![\blue\star](https://tex.z-dn.net/?f=%5Cblue%5Cstar)
Given</h3>
- mass = 200kg
- acceleration = 10m\s^2
Star coding
<h3>
![\blue\star](https://tex.z-dn.net/?f=%5Cblue%5Cstar)
As we know that force = mass × acceleration</h3>
- F = m×a
- F = 200 ×10
- F = 2000N
<h3>
![\blue\star](https://tex.z-dn.net/?f=%5Cblue%5Cstar)
so mate here is ur ans force is equal to 2000N.</h3>
Hope it helps
Answer:
The larger constant goes to the spring with the largest F/x ratio, this is by measuring a larger Force or by displacing less from the equilibrium position
Explanation:
Using Hooke's Law we have that for each spring the following relations is fulfilled:
![F=-kx](https://tex.z-dn.net/?f=F%3D-kx)
Where F is the restoring force (opposite to displacement ergo the negative sign), k is the constant and x is the displacement.
All we need to do is to compare the following:
![k=\frac{F}{x}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7BF%7D%7Bx%7D)
The larger constant goes to the spring with the largest F/x ratio, this is by measuring a larger Force or by displacing less from the equilibrium position
If you have numerical data we can run the calculations