This is the same question as the one previously but with more details, so I will just use my previous answer.
1800 to 1820 is 20 minutes.1830 to 1838 is 8 minutes.1840 to 1905 is 25 minutes.
The total time travelled is 20+8+25 = 53 minutes = 3180 seconds.
The distance between Glasgow and Edinburgh is 28 + 12 + 34 = 74 km = 74000 m.
So, the average speed is 74000m/3180s = 23.27 m/s (4 s.f.)
The orbital radius is: ![r=\frac{GM}{v^2}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7BGM%7D%7Bv%5E2%7D)
Explanation:
The problem is asking to find the radius of the orbit of a satellite around a planet, given the orbital speed of the satellite.
For a satellite in orbit around a planet, the gravitational force provides the required centripetal force to keep it in circular motion, therefore we can write:
![\frac{GMm}{r^2}=m\frac{v^2}{r}](https://tex.z-dn.net/?f=%5Cfrac%7BGMm%7D%7Br%5E2%7D%3Dm%5Cfrac%7Bv%5E2%7D%7Br%7D)
where
G is the gravitational constant
M is the mass of the planet
m is the mass of the satellite
r is the radius of the orbit
v is the speed of the satellite
Re-arranging the equation, we find:
![\frac{GM}{r}=v^2\\r=\frac{GM}{v^2}](https://tex.z-dn.net/?f=%5Cfrac%7BGM%7D%7Br%7D%3Dv%5E2%5C%5Cr%3D%5Cfrac%7BGM%7D%7Bv%5E2%7D)
Learn more about circular motion:
brainly.com/question/2562955
brainly.com/question/6372960
#LearnwithBrainly
<h2>Answer:</h2><h3>(A) the positively charged surface increases and the energy stored in the capacitor increases.</h3>
When charging a capacitor transferring charge from one surface to the other, the first surface becomes negatively charged while the second surface becomes positively charged. As you transfer the charge, the voltage of the positively charged surface increases and the energy stored in the capacitor also increases. We can solve this by the definition of <em>capacitance</em><em> </em>that is <em>a measure of the ability of a capacitor to store energy. </em>For any capacitor, the capacitance is a constant defined as:
![C=\frac{Q}{V_{ab}}](https://tex.z-dn.net/?f=C%3D%5Cfrac%7BQ%7D%7BV_%7Bab%7D%7D)
To maintain
constant, if Q increases V also increases.
On the other hand, the potential energy
can be expressed as:
![U=\frac{Q^{2}}{2C}](https://tex.z-dn.net/?f=U%3D%5Cfrac%7BQ%5E%7B2%7D%7D%7B2C%7D)
In conclusion, as Q increases the potential energy also increases.