Answer:
ω2 = 216.47 rad/s
Explanation:
given data
radius r1 = 460 mm
radius r2 = 46 mm
ω = 32k rad/s
solution
we know here that power generated by roller that is
power = T. ω ..............1
power = F × r × ω
and this force of roller on cylinder is equal and opposite force apply by roller
so power transfer equal in every cylinder so
( F × r1 × ω1) ÷ 2 = ( F × r2 × ω2 ) ÷ 2 ................2
so
ω2 =
ω2 = 216.47
Answer:
Magnitude of vector A = 0.904
Explanation:
Vector A , which is directed along an x axis, that is
![\vec{A}=x_A\hat{i}](https://tex.z-dn.net/?f=%5Cvec%7BA%7D%3Dx_A%5Chat%7Bi%7D)
Vector B , which has a magnitude of 5.5 m
![\vec{B}=x_B\hat{i}+y_B\hat{j}](https://tex.z-dn.net/?f=%5Cvec%7BB%7D%3Dx_B%5Chat%7Bi%7D%2By_B%5Chat%7Bj%7D)
![\sqrt{x_{B}^{2}+y_{B}^{2}}=5.5\\\\x_{B}^{2}+y_{B}^{2}=30.25](https://tex.z-dn.net/?f=%5Csqrt%7Bx_%7BB%7D%5E%7B2%7D%2By_%7BB%7D%5E%7B2%7D%7D%3D5.5%5C%5C%5C%5Cx_%7BB%7D%5E%7B2%7D%2By_%7BB%7D%5E%7B2%7D%3D30.25)
The sum is a third vector that is directed along the y axis, with a magnitude that is 6.0 times that of vector A ![\vec{A}+\vec{B}=6x_A\hat{j}\\\\x_A\hat{i}+x_B\hat{i}+y_B\hat{j}=6x_A\hat{j}](https://tex.z-dn.net/?f=%5Cvec%7BA%7D%2B%5Cvec%7BB%7D%3D6x_A%5Chat%7Bj%7D%5C%5C%5C%5Cx_A%5Chat%7Bi%7D%2Bx_B%5Chat%7Bi%7D%2By_B%5Chat%7Bj%7D%3D6x_A%5Chat%7Bj%7D)
Comparing we will get
![x_A=-x_B\\\\y_B=6x_A](https://tex.z-dn.net/?f=x_A%3D-x_B%5C%5C%5C%5Cy_B%3D6x_A)
Substituting in ![x_{B}^{2}+y_{B}^{2}=30.25](https://tex.z-dn.net/?f=x_%7BB%7D%5E%7B2%7D%2By_%7BB%7D%5E%7B2%7D%3D30.25)
![\left (-x_{A} \right )^{2}+\left (6x_{A} \right )^{2}=30.25\\\\37x_{A}^2=30.25\\\\x_{A}=0.904](https://tex.z-dn.net/?f=%5Cleft%20%28-x_%7BA%7D%20%5Cright%20%29%5E%7B2%7D%2B%5Cleft%20%286x_%7BA%7D%20%5Cright%20%29%5E%7B2%7D%3D30.25%5C%5C%5C%5C37x_%7BA%7D%5E2%3D30.25%5C%5C%5C%5Cx_%7BA%7D%3D0.904)
So we have
![\vec{A}=0.904\hat{i}](https://tex.z-dn.net/?f=%5Cvec%7BA%7D%3D0.904%5Chat%7Bi%7D)
Magnitude of vector A = 0.904
Answer:
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