Answer:
this is what i always used
Explanation:
http://www.sjutsscience.com/uploads/3/7/4/5/37458459/phet_collision_lab_2020_key.pdf
This specific question applies the concept of relative velocity. Relative velocity is describes as the velocity of an object with respect to another object, whether static or moving. For this problem, the correct statement is t<span>he spotter above the tunnel will observe the speed of the spy as 95 km/h, and the person in the car will observe the speed of the spy as 200 km/h. The following are the justification, relative speed of the spotter at the tunnel is equal to velocity (Vss) = velocity spy relative to the ground (Vsg) + V ground relative to the spotter (Vgs). Thus, V = 95 + 0 km/h = 95 km/h. Note that the sign of velocity is dependent on the direction, hence, opposite direction is equal to negative velocity. For the second condition, velocity of the spy in reference person in the car is equal to velocity of the spy reference to ground plus the velocity of the ground reference to the person in the car. Thus, V = 95 km/h + 105 km/h = 200 km/h. Please be cautious in assigning the sign of the velocity.</span>
The answer is 15.0N its explained in newtons third law hope this helps:)
If there is no friction, then NO force is needed to keep an object moving. Go back and look at Newton's first law of motion again.
Answer:
a) ![T_{min} = 80\,s](https://tex.z-dn.net/?f=T_%7Bmin%7D%20%3D%2080%5C%2Cs)
Explanation:
a) Let consider that disk accelerates and decelerates at constant rate. The expression for angular acceleration and deceleration are, respectively:
Acceleration
![\alpha_{1} = \frac{\omega_{max}^{2}}{2\cdot (400\,rad)}](https://tex.z-dn.net/?f=%5Calpha_%7B1%7D%20%3D%20%5Cfrac%7B%5Comega_%7Bmax%7D%5E%7B2%7D%7D%7B2%5Ccdot%20%28400%5C%2Crad%29%7D)
Deceleration
![\alpha_{2} = -\frac{\omega_{max}^{2}}{2\cdot (400\,rad)}](https://tex.z-dn.net/?f=%5Calpha_%7B2%7D%20%3D%20-%5Cfrac%7B%5Comega_%7Bmax%7D%5E%7B2%7D%7D%7B2%5Ccdot%20%28400%5C%2Crad%29%7D)
Since angular acceleration and deceleration have same magnitude but opposite sign. Let is find the maximum allowed angular speed from maximum allowed centripetal acceleration:
![a_{r,max} = \omega_{max}^{2}\cdot r](https://tex.z-dn.net/?f=a_%7Br%2Cmax%7D%20%3D%20%5Comega_%7Bmax%7D%5E%7B2%7D%5Ccdot%20r)
![\omega_{max} = \sqrt{\frac{a_{r,max}}{r} }](https://tex.z-dn.net/?f=%5Comega_%7Bmax%7D%20%3D%20%5Csqrt%7B%5Cfrac%7Ba_%7Br%2Cmax%7D%7D%7Br%7D%20%7D)
![\omega_{max} = \sqrt{\frac{100\,\frac{m}{s^{2}} }{0.25\,m} }](https://tex.z-dn.net/?f=%5Comega_%7Bmax%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B100%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%7D%7B0.25%5C%2Cm%7D%20%7D)
![\omega_{max} = 20\,\frac{rad}{s}](https://tex.z-dn.net/?f=%5Comega_%7Bmax%7D%20%3D%2020%5C%2C%5Cfrac%7Brad%7D%7Bs%7D)
Maximum magnitude of acceleration/deceleration is:
![\alpha = 0.5\,\frac{rad}{s^{2}}](https://tex.z-dn.net/?f=%5Calpha%20%3D%200.5%5C%2C%5Cfrac%7Brad%7D%7Bs%5E%7B2%7D%7D)
The least time require for rotation is:
![T_{min} = 2\cdot \left(\frac{20\,\frac{rad}{s} }{0.5\,\frac{rad}{s^{2}} } \right)](https://tex.z-dn.net/?f=T_%7Bmin%7D%20%3D%202%5Ccdot%20%5Cleft%28%5Cfrac%7B20%5C%2C%5Cfrac%7Brad%7D%7Bs%7D%20%7D%7B0.5%5C%2C%5Cfrac%7Brad%7D%7Bs%5E%7B2%7D%7D%20%7D%20%20%5Cright%29)
![T_{min} = 80\,s](https://tex.z-dn.net/?f=T_%7Bmin%7D%20%3D%2080%5C%2Cs)