Answer:
A
Explanation:
The answer would be A because when feet push down on anything, the force will push down the skateboard
Answer:
![T_{max} = 4.735\,kN\cdot m](https://tex.z-dn.net/?f=T_%7Bmax%7D%20%3D%204.735%5C%2CkN%5Ccdot%20m)
Explanation:
The shear stress due to torque can be calculed by using the following model:
![\tau_{max} = \frac{T_{max}\cdot r_{ext}}{J_{tube}}](https://tex.z-dn.net/?f=%5Ctau_%7Bmax%7D%20%3D%20%5Cfrac%7BT_%7Bmax%7D%5Ccdot%20r_%7Bext%7D%7D%7BJ_%7Btube%7D%7D)
The maximum torque on the section is:
![T_{max} = \frac{\tau_{max}\cdot J_{tube}}{r_{ext}}](https://tex.z-dn.net/?f=T_%7Bmax%7D%20%3D%20%5Cfrac%7B%5Ctau_%7Bmax%7D%5Ccdot%20J_%7Btube%7D%7D%7Br_%7Bext%7D%7D)
The Torsion Constant for the circular tube is:
![J_{tube} = \frac{\pi}{32}\cdot (D_{ext}^{4}-D_{int}^{4})](https://tex.z-dn.net/?f=J_%7Btube%7D%20%3D%20%5Cfrac%7B%5Cpi%7D%7B32%7D%5Ccdot%20%28D_%7Bext%7D%5E%7B4%7D-D_%7Bint%7D%5E%7B4%7D%29)
![J_{tube} = \frac{\pi}{4}\cdot [(0.053\,m)^{4}-(0.038\,m)^{4}]](https://tex.z-dn.net/?f=J_%7Btube%7D%20%3D%20%5Cfrac%7B%5Cpi%7D%7B4%7D%5Ccdot%20%5B%280.053%5C%2Cm%29%5E%7B4%7D-%280.038%5C%2Cm%29%5E%7B4%7D%5D)
![J_{tube} = 4.560\times 10^{-6}\,m^{4}](https://tex.z-dn.net/?f=J_%7Btube%7D%20%3D%204.560%5Ctimes%2010%5E%7B-6%7D%5C%2Cm%5E%7B4%7D)
Now, the require output is computed:
![T_{max} = \frac{(27\times 10^{3}\,kPa)\cdot (4.560\times 10^{-6}\,m^{4})}{0.026\,m}](https://tex.z-dn.net/?f=T_%7Bmax%7D%20%3D%20%5Cfrac%7B%2827%5Ctimes%2010%5E%7B3%7D%5C%2CkPa%29%5Ccdot%20%284.560%5Ctimes%2010%5E%7B-6%7D%5C%2Cm%5E%7B4%7D%29%7D%7B0.026%5C%2Cm%7D)
![T_{max} = 4.735\,kN\cdot m](https://tex.z-dn.net/?f=T_%7Bmax%7D%20%3D%204.735%5C%2CkN%5Ccdot%20m)
Increasing the level of an incline:
Increases final velocity
Increases the work done
Increases the initial potential energy
Increases the final kinetic energy
Answer:
The speed of the top of the wheel is twice the speed of the car.
That is: 72 m/s
Explanation:
To find the speed of the top of the wheel, we need to combine to velocities: the tangential velocity of the rotating wheel due to rotational motion
- with
being the wheel's angular velocity,
plus the velocity due to the translation of the center of mass (v = 36 m/s).
The wheel's angular velocity (in radians per second) can be obtained using the tangential velocity for the pure rotational motion and it equals:![\omega=\frac{v_t}{r} =\frac{36}{0.25} \,s^{-1}](https://tex.z-dn.net/?f=%5Comega%3D%5Cfrac%7Bv_t%7D%7Br%7D%20%3D%5Cfrac%7B36%7D%7B0.25%7D%20%5C%2Cs%5E%7B-1%7D)
Then the addition of these two velocities equals:
![\omega\,R+v=\frac{36}{0.25} (0.25)\,\,\frac{m}{s} +36\,\,\frac{m}{s} =72\,\,\frac{m}{s}](https://tex.z-dn.net/?f=%5Comega%5C%2CR%2Bv%3D%5Cfrac%7B36%7D%7B0.25%7D%20%280.25%29%5C%2C%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%2B36%5C%2C%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%3D72%5C%2C%5C%2C%5Cfrac%7Bm%7D%7Bs%7D)
This problem is relatively simple, in order to solve this problem the only formula you need to know is the formula for friction, which is:
<span>Ff = UsN </span>
<span>where Us is the coefficient of static friction and N is the normal force. </span>
<span>In order to get the crate moving you must first apply enough force to overcome the static friction: </span>
<span>Fapplied = Ff </span>
<span>Since Fapplied = 43 Newtons: </span>
<span>Fapplied = Ff = 43 = UsN </span>
<span>and it was given that Us = 0.11, so all you have to do is isolate N by dividing both sides by 0.11 </span>
<span>43/0.11 = N = 390.9 which is approximately 391 or C. 3.9x10^2</span>