Distance is the total length of an object's path. Displacement is the overall change in position, ie. how far an object is from its initial position.
The court is 30 m long, so a path going back and forth once is 60 m long. Going along this path 6 times totals 360 m.
The end point is the same as the starting point, so the displacement is 0 m.
Answer:
For cast iron we have
![h = 0.92 m](https://tex.z-dn.net/?f=h%20%3D%200.92%20m)
For copper
![h = 1.05 m](https://tex.z-dn.net/?f=h%20%3D%201.05%20m)
For Lead
![h = 1.23 m](https://tex.z-dn.net/?f=h%20%3D%201.23%20m)
For Zinc
![h = 2.43 m](https://tex.z-dn.net/?f=h%20%3D%202.43%20m)
Explanation:
As we know that final speed of the block is calculated by work energy theorem
![W_f + W_g = \frac{1}{2}mv^2](https://tex.z-dn.net/?f=W_f%20%2B%20W_g%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
now we have
![-\mu_k mg cos\theta(\frac{h}{sin\theta}) + mgh = \frac{1}{2}mv^2](https://tex.z-dn.net/?f=-%5Cmu_k%20mg%20cos%5Ctheta%28%5Cfrac%7Bh%7D%7Bsin%5Ctheta%7D%29%20%2B%20mgh%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
now we have
![v^2 = 2gh - 2\mu_k g h cot\theta](https://tex.z-dn.net/?f=v%5E2%20%3D%202gh%20-%202%5Cmu_k%20g%20h%20cot%5Ctheta)
![v = \sqrt{2gh(1 - \mu_k cot\theta)}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B2gh%281%20-%20%5Cmu_k%20cot%5Ctheta%29%7D)
For cast iron we have
![4 = \sqrt{2(9.81)(h)(1 - 0.15cot52)}](https://tex.z-dn.net/?f=4%20%3D%20%5Csqrt%7B2%289.81%29%28h%29%281%20-%200.15cot52%29%7D)
![h = 0.92 m](https://tex.z-dn.net/?f=h%20%3D%200.92%20m)
For copper
![4 = \sqrt{2(9.81)(h)(1 - 0.29cot52)}](https://tex.z-dn.net/?f=4%20%3D%20%5Csqrt%7B2%289.81%29%28h%29%281%20-%200.29cot52%29%7D)
![h = 1.05 m](https://tex.z-dn.net/?f=h%20%3D%201.05%20m)
For Lead
![4 = \sqrt{2(9.81)(h)(1 - 0.43cot52)}](https://tex.z-dn.net/?f=4%20%3D%20%5Csqrt%7B2%289.81%29%28h%29%281%20-%200.43cot52%29%7D)
![h = 1.23 m](https://tex.z-dn.net/?f=h%20%3D%201.23%20m)
For Zinc
![4 = \sqrt{2(9.81)(h)(1 - 0.85cot52)}](https://tex.z-dn.net/?f=4%20%3D%20%5Csqrt%7B2%289.81%29%28h%29%281%20-%200.85cot52%29%7D)
![h = 2.43 m](https://tex.z-dn.net/?f=h%20%3D%202.43%20m)
It is equal to the force you exert on the chair.
The color should be gray, tin
Answer:
![v=6.65m/sec](https://tex.z-dn.net/?f=v%3D6.65m%2Fsec)
Explanation:
From the Question we are told that:
Mass ![m=97.6](https://tex.z-dn.net/?f=m%3D97.6)
Coefficient of kinetic friction ![\mu k=0.555](https://tex.z-dn.net/?f=%5Cmu%20k%3D0.555)
Generally the equation for Frictional force is mathematically given by
![F=\mu mg](https://tex.z-dn.net/?f=F%3D%5Cmu%20mg)
![F=0.555*97.6*9.8](https://tex.z-dn.net/?f=F%3D0.555%2A97.6%2A9.8)
![F=531.388N](https://tex.z-dn.net/?f=F%3D531.388N)
Generally the Newton's equation for Acceleration due to Friction force is mathematically given by
![a_f=-\mu g](https://tex.z-dn.net/?f=a_f%3D-%5Cmu%20g)
![a_f=-0.555 *9.81](https://tex.z-dn.net/?f=a_f%3D-0.555%20%2A9.81)
![a_f=-54455m/sec^2](https://tex.z-dn.net/?f=a_f%3D-54455m%2Fsec%5E2)
Therefore
![v=u-at](https://tex.z-dn.net/?f=v%3Du-at)
![v=0+5.45*1.22](https://tex.z-dn.net/?f=v%3D0%2B5.45%2A1.22)
![v=6.65m/sec](https://tex.z-dn.net/?f=v%3D6.65m%2Fsec)