The acceleration due to gravity of the planet X is 1 m/s².
The given parameters;
- height above the ground, h = 100 m
- initial velocity of the rock, u = 15 m/s
- time of motion of the rock, t = 10 s
The acceleration due to gravity is calculated as follows;

Thus, the acceleration due to gravity of the planet X is 1 m/s²
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The ball's gravitational potential energy is converted into kinetic energy as it falls toward the ground.
<h3>How can the height of a dropped ball be determined?</h3>
Y = 1/2 g t 2, where y is the height above the ground, g = 9.8 m/s2, and t = 1.3 s, is the formula for problems like these. Any freely falling body with an initial velocity of zero meters per second can use this formula. figuring out how much y is.
A ball drops from the top of a building and picks up speed as it descends. Its speed is increasing by 10 m/s every second. What we refer to as motion with constant acceleration is, for example, a ball falling due to gravity.
The ball's parabolic motion causes it to move at a speed of 26.3 m/s right before it strikes the ground, which is faster than its straight downhill motion, which has a speed of 17.1 m/s. Take note of the rising positive y direction in the above graphic.
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Answer:
19.2 m/s
Explanation:
The train is moving at 18 m/s and you are walking in the same direction (east) so the speeds are added
18 + 1.2 = 19.2
If you were walking backwards (west) your velocity with respect to the ground would be
18 - 1.2 = 16.8
Answer:
The value is 
Explanation:
From the question we are told that
The radius of the inner conductor is 
The radius of the outer conductor is 
The potential at the outer conductor is 
Generally the capacitance per length of the capacitor like set up of the two conductors is
![C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}](https://tex.z-dn.net/?f=C%3D%20%5Cfrac%7B2%20%2A%20%5Cpi%20%2A%20%5Cepsilon_o%20%7D%7B%20ln%20%5B%5Cfrac%7Br_2%7D%7Br_1%7D%20%5D%7D)
Here
is the permitivity of free space with value 
=> ![C= \frac{2 * 3.142 * 8.85*10^{-12} }{ ln [\frac{0.003}{0.001} ]}](https://tex.z-dn.net/?f=C%3D%20%5Cfrac%7B2%20%2A%20%203.142%20%20%2A%208.85%2A10%5E%7B-12%7D%20%20%7D%7B%20ln%20%5B%5Cfrac%7B0.003%7D%7B0.001%7D%20%5D%7D)
=> 
Generally given that the potential of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge
Generally the line charge density of the outer conductor is mathematically represented as

=> 
=> 
Generally the surface charge density is mathematically represented as
here 
=> 
=> 