The nucleus is always positive since thats where the protons are located and the neutrons have no charge....
Given:
Total distance = 1000 kilometer
Total time = 5 hours
To find:
Average speed = ?
Formula used:
= ![\frac{s}{t}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bs%7D%7Bt%7D%20%20)
Where
= average speed
s = total distance
t = total time
Solution:
Average speed of the jet is given by,
= ![\frac{s}{t}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bs%7D%7Bt%7D%20%20)
Where
= average speed
s = total distance
t = total time
= ![\frac{1000}{5}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1000%7D%7B5%7D%20%20)
= 200 km/ h
Thus, average speed of the jet is 200 km/h.
Hence, Option (A) is correct.
Huh? pls elaborate i don’t understand
To solve this problem, apply the concepts related to the centripetal acceleration as the equivalent of gravity, and the kinematic equations of linear motion that will relate the speed, the distance traveled and the period of the body to meet the needs given in the problem. Centripetal acceleration is defined as,
![a_c = \frac{v^2}{r}](https://tex.z-dn.net/?f=a_c%20%3D%20%5Cfrac%7Bv%5E2%7D%7Br%7D)
Here,
v = Tangential Velocity
r = Radius
If we rearrange the equation to get the velocity we have,
![v = \sqrt{a_c r}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7Ba_c%20r%7D)
But at this case the centripetal acceleration must be equal to the gravitational at the Earth, then
![v = \sqrt{gr}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7Bgr%7D)
![v = \sqrt{(9.8)(\frac{1000}{2})}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%289.8%29%28%5Cfrac%7B1000%7D%7B2%7D%29%7D)
![v = 70m/s](https://tex.z-dn.net/?f=v%20%3D%2070m%2Fs)
The perimeter of the cylinder would be given by,
![\phi = 2\pi r](https://tex.z-dn.net/?f=%5Cphi%20%3D%202%5Cpi%20r)
![\phi = 2\pi (500m)](https://tex.z-dn.net/?f=%5Cphi%20%3D%202%5Cpi%20%28500m%29)
![\phi = 3141.6m](https://tex.z-dn.net/?f=%5Cphi%20%3D%203141.6m)
Therefore now related by kinematic equations of linear motion the speed with the distance traveled and the time we will have to
![v = \frac{d}{t} \rightarrow \text{ But here } d = \phi](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bd%7D%7Bt%7D%20%5Crightarrow%20%5Ctext%7B%20But%20here%20%7D%20d%20%3D%20%5Cphi)
![v = \frac{\phi}{t} \rightarrow t = \frac{\phi}{t}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B%5Cphi%7D%7Bt%7D%20%5Crightarrow%20t%20%3D%20%5Cfrac%7B%5Cphi%7D%7Bt%7D)
![t = \frac{3141.6}{70m/s}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B3141.6%7D%7B70m%2Fs%7D)
![t = 44.9s](https://tex.z-dn.net/?f=t%20%3D%2044.9s)
Therefore the period will be 44.9s