Answer:
The average acceleration of the bearings is ![0.77\times10^{3}\ m/s^2](https://tex.z-dn.net/?f=0.77%5Ctimes10%5E%7B3%7D%5C%20m%2Fs%5E2)
Explanation:
Given that,
Height = 1.94 m
Bounced height = 1.48 m
Time interval ![t=14.86\times10^{-3}\ s](https://tex.z-dn.net/?f=t%3D14.86%5Ctimes10%5E%7B-3%7D%5C%20s)
Velocity of the ball bearing just before hitting the steel plate
We need to calculate the velocity
Using conservation of energy
![mgh=\dfrac{1}{2}mv^2](https://tex.z-dn.net/?f=mgh%3D%5Cdfrac%7B1%7D%7B2%7Dmv%5E2)
Put the value into the formula
![9.8\times1.94=\dfrac{1}{2}\times v^2](https://tex.z-dn.net/?f=9.8%5Ctimes1.94%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20v%5E2)
![v=\sqrt{2\times9.8\times1.94}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B2%5Ctimes9.8%5Ctimes1.94%7D)
![v=6.166\ m/s](https://tex.z-dn.net/?f=v%3D6.166%5C%20m%2Fs)
Negative as it is directed downwards
After bounce back,
We need to calculate the velocity
Using conservation of energy
![mgh=\dfrac{1}{2}mv^2'](https://tex.z-dn.net/?f=mgh%3D%5Cdfrac%7B1%7D%7B2%7Dmv%5E2%27)
Put the value into the formula
![9.8\times1.48=\dfrac{1}{2}\times v^2'](https://tex.z-dn.net/?f=9.8%5Ctimes1.48%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20v%5E2%27)
![v'=\sqrt{2\times9.8\times1.48}](https://tex.z-dn.net/?f=v%27%3D%5Csqrt%7B2%5Ctimes9.8%5Ctimes1.48%7D)
![v'=5.38\ m/s](https://tex.z-dn.net/?f=v%27%3D5.38%5C%20m%2Fs)
We need to calculate the average acceleration of the bearings while they are in contact with the plate
Using formula of acceleration
![a=\dfrac{v-v'}{t}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bv-v%27%7D%7Bt%7D)
Put the value into the formula
![a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B5.38-%28-6.166%29%7D%7B14.86%5Ctimes10%5E%7B-3%7D%7D)
![a=776.98\ m/s^2](https://tex.z-dn.net/?f=a%3D776.98%5C%20m%2Fs%5E2)
![a=0.77\times10^{3}\ m/s^2](https://tex.z-dn.net/?f=a%3D0.77%5Ctimes10%5E%7B3%7D%5C%20m%2Fs%5E2)
Hence,The average acceleration of the bearings is ![0.77\times10^{3}\ m/s^2](https://tex.z-dn.net/?f=0.77%5Ctimes10%5E%7B3%7D%5C%20m%2Fs%5E2)
Answer:
The non-relativistic kinetic energy of a proton is ![6.76\times10^{-12}\ J](https://tex.z-dn.net/?f=6.76%5Ctimes10%5E%7B-12%7D%5C%20J)
The relativistic kinetic energy of a proton is ![7.25\times10^{-12}\ m/s](https://tex.z-dn.net/?f=7.25%5Ctimes10%5E%7B-12%7D%5C%20m%2Fs)
Explanation:
Given that,
Mass of proton ![m=1.67\times10^{-27}\ kg](https://tex.z-dn.net/?f=m%3D1.67%5Ctimes10%5E%7B-27%7D%5C%20kg)
Speed
We need to calculate the kinetic energy for non relativistic
Using formula of kinetic energy
![K.E=\dfrac{1}{2}mv^2](https://tex.z-dn.net/?f=K.E%3D%5Cdfrac%7B1%7D%7B2%7Dmv%5E2)
Put the value into the formula
![K.E=\dfrac{1}{2}\times1.67\times10^{-27}\times(9.00\times10^{7})^2](https://tex.z-dn.net/?f=K.E%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes1.67%5Ctimes10%5E%7B-27%7D%5Ctimes%289.00%5Ctimes10%5E%7B7%7D%29%5E2)
![K.E=6.76\times10^{-12}\ J](https://tex.z-dn.net/?f=K.E%3D6.76%5Ctimes10%5E%7B-12%7D%5C%20J)
We need to calculate the kinetic energy for relativistic
Using formula of kinetic energy
![K.E=mc^2(\sqrt{(\dfrac{1}{1-\dfrac{v^2}{c^2}})}-1)](https://tex.z-dn.net/?f=K.E%3Dmc%5E2%28%5Csqrt%7B%28%5Cdfrac%7B1%7D%7B1-%5Cdfrac%7Bv%5E2%7D%7Bc%5E2%7D%7D%29%7D-1%29)
![K.E=1.67\times10^{-27}\times(3\times10^{8})^{2}\cdot\left(\sqrt{\frac{1}{1-\frac{\left(9.00\times10^{7}\right)^{2}}{(3\times10^{8})^{2}}}}-1\right)](https://tex.z-dn.net/?f=K.E%3D1.67%5Ctimes10%5E%7B-27%7D%5Ctimes%283%5Ctimes10%5E%7B8%7D%29%5E%7B2%7D%5Ccdot%5Cleft%28%5Csqrt%7B%5Cfrac%7B1%7D%7B1-%5Cfrac%7B%5Cleft%289.00%5Ctimes10%5E%7B7%7D%5Cright%29%5E%7B2%7D%7D%7B%283%5Ctimes10%5E%7B8%7D%29%5E%7B2%7D%7D%7D%7D-1%5Cright%29)
![K.E=7.25\times10^{-12}\ m/s](https://tex.z-dn.net/?f=K.E%3D7.25%5Ctimes10%5E%7B-12%7D%5C%20m%2Fs)
Hence, The non-relativistic kinetic energy of a proton is ![6.76\times10^{-12}\ J](https://tex.z-dn.net/?f=6.76%5Ctimes10%5E%7B-12%7D%5C%20J)
The relativistic kinetic energy of a proton is ![7.25\times10^{-12}\ m/s](https://tex.z-dn.net/?f=7.25%5Ctimes10%5E%7B-12%7D%5C%20m%2Fs)
Answer: Instantaneous speed.
Explanation:
Answer:
A driver.
Explanation:
Using a driver while at least 350 yds away is better than using a iron, because it will be a waste of the par 4 as it is not as powerful as the driver.