Answer:
a).![x_{t}=583.3rad](https://tex.z-dn.net/?f=x_%7Bt%7D%3D583.3rad)
b).![t_{total}=10.52s](https://tex.z-dn.net/?f=t_%7Btotal%7D%3D10.52s)
c).![a=12.62 \frac{rad}{s^2}](https://tex.z-dn.net/?f=a%3D12.62%20%5Cfrac%7Brad%7D%7Bs%5E2%7D)
Explanation:
The angular acceleration is constant so we can use the formulas of uniform motion with the model of angular acceleration
a).
![x_{r}=x_{i}+v_{i}+\frac{1}{2}a_{a}*t^2](https://tex.z-dn.net/?f=x_%7Br%7D%3Dx_%7Bi%7D%2Bv_%7Bi%7D%2B%5Cfrac%7B1%7D%7B2%7Da_%7Ba%7D%2At%5E2)
![x_{r}=0+28.0\frac{rad}{s}*2.20s+\frac{1}{2}*35.0\frac{rad}{s^2}*2.20s](https://tex.z-dn.net/?f=x_%7Br%7D%3D0%2B28.0%5Cfrac%7Brad%7D%7Bs%7D%2A2.20s%2B%5Cfrac%7B1%7D%7B2%7D%2A35.0%5Cfrac%7Brad%7D%7Bs%5E2%7D%2A2.20s)
![x_{r}=146.3rad](https://tex.z-dn.net/?f=x_%7Br%7D%3D146.3rad)
so the total angle between t=0 and the time it stopped is
![x_{t}=146.3rad+437rad=583.3rad](https://tex.z-dn.net/?f=x_%7Bt%7D%3D146.3rad%2B437rad%3D583.3rad)
b).
![w_{f}=w_{o}+a*t](https://tex.z-dn.net/?f=w_%7Bf%7D%3Dw_%7Bo%7D%2Ba%2At)
=![w_{o}](https://tex.z-dn.net/?f=w_%7Bo%7D)
![x_{t}-x_{r}=\frac{1}{2}*(w_{o}-w_{f})*t](https://tex.z-dn.net/?f=x_%7Bt%7D-x_%7Br%7D%3D%5Cfrac%7B1%7D%7B2%7D%2A%28w_%7Bo%7D-w_%7Bf%7D%29%2At)
![583.3-146.3=\frac{1}{2}*(105-0)*t](https://tex.z-dn.net/?f=583.3-146.3%3D%5Cfrac%7B1%7D%7B2%7D%2A%28105-0%29%2At)
![t=\frac{437rad}{105\frac{rad}{s}}=8.32s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B437rad%7D%7B105%5Cfrac%7Brad%7D%7Bs%7D%7D%3D8.32s)
![t_{total}=8.32+2.2=10.52s](https://tex.z-dn.net/?f=t_%7Btotal%7D%3D8.32%2B2.2%3D10.52s)
c).
![w_{f}=w_{o}+a*t](https://tex.z-dn.net/?f=w_%7Bf%7D%3Dw_%7Bo%7D%2Ba%2At)
![0=105 rad/s+a*8.32s](https://tex.z-dn.net/?f=0%3D105%20rad%2Fs%2Ba%2A8.32s)
![a=\frac{105 rad/s}{8.32s}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B105%20rad%2Fs%7D%7B8.32s%7D)
![a=12.62 \frac{rad}{s^2}](https://tex.z-dn.net/?f=a%3D12.62%20%5Cfrac%7Brad%7D%7Bs%5E2%7D)
Answer:
T = 300 N
Explanation:
As we know that the painter is standing at the middle of the scaffold
So here the two ropes will exert same tension on the scaffold
So we will have
![T_1 + T_2 = W](https://tex.z-dn.net/?f=T_1%20%2B%20T_2%20%3D%20W)
also we know
![T_1 = T_2](https://tex.z-dn.net/?f=T_1%20%3D%20T_2)
so we will have
![2T = W](https://tex.z-dn.net/?f=2T%20%3D%20W)
![T = \frac{W}{2}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7BW%7D%7B2%7D)
so tension is half of the weight in both the strings
So it is
T = 300 N
Answer:
efficiancy=40 percent
Explanation:
efficiency=energy output/energy input×100
efficiancy=8J/20J×100
efficiancy=0.4×100
efficiancy=40 percent
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