Definitely not the last 2. My bet is on the first option. If it is wrong don't hit me please...
Answer:
B. w=12.68rad/s
C. α=3.52rad/s^2
Explanation:
B)
We can solve this problem by taking into account that (as in the uniformly accelerated motion)
( 1 )
where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.
In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have
![\theta=\frac{3.7m}{0.162m}=22.83rad](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cfrac%7B3.7m%7D%7B0.162m%7D%3D22.83rad)
![22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}](https://tex.z-dn.net/?f=22.83rad%3D%5Cfrac%7B1%7D%7B2%7D%5Calpha%20%283.6s%29%5E2%5C%5C%5C%5C%5Calpha%3D2%5Cfrac%7B%2822.83rad%29%7D%7B3.6%5E2s%7D%3D3.52%5Cfrac%7Brad%7D%7Bs%5E2%7D)
to calculate the angular speed w we can use![\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}](https://tex.z-dn.net/?f=%5Calpha%3D%5Cfrac%7B%5Comega%20_%7Bf%7D-%5Comega%20_%7Bi%7D%7D%7Bt%20_%7Bf%7D-t%20_%7Bi%7D%7D%5C%5C%5C%5C%5Comega_%7Bf%7D%3D%5Calpha%20t_%7Bf%7D%3D%283.52%5Cfrac%7Brad%7D%7Bs%5E2%7D%29%283.6%29%3D12.68%5Cfrac%7Brad%7D%7Bs%7D)
Thus, wf=12.68rad/s
C) We can use our result in B)
![\alpha=3.52\frac{rad}{s^2}](https://tex.z-dn.net/?f=%5Calpha%3D3.52%5Cfrac%7Brad%7D%7Bs%5E2%7D)
I hope this is useful for you
regards
Answer:
The component of the force due to gravity perpendicular and parallel to the slope is 113.4 N and 277.8 N respectively.
Explanation:
Force is any cause capable of modifying the state of motion or rest of a body or of producing a deformation in it. Any force can be decomposed into two vectors, so that the sum of both vectors matches the vector before decomposing. The decomposition of a force into its components can be done in any direction.
Taking into account the simple trigonometric relations, such as sine, cosine and tangent, the value of their components and the value of the angle of application, then the parallel and perpendicular components will be:
- Fparallel = F*sinα =300 N*sin 67.8° =300 N*0.926⇒ Fparallel =277.8 N
- Fperpendicular = F*cosα = 300 N*cos 67.8° = 300 N*0.378 ⇒ Fperpendicular= 113.4 N
<u><em>The component of the force due to gravity perpendicular and parallel to the slope is 113.4 N and 277.8 N respectively.</em></u>
Washington DC and new Mexico