Eₖ = √1/2 mv²
Eₖ/v² = 1/2 m
2Eₖ/v² = m
mgh = 1/2 mv²
gh = 1/2 v²
2gh = v²
v = √2gh
= √2 × 10 N/kg × 2 m
= 6.324 ms⁻¹
m = 9.8/(6424)²
= 9.8/39.993
= 0.245 g
I am not sure if it's correct.
i transposed for m in the first equation then to find the velocity I combined the potential and kinetic energy equation. there was no mass because velocity doesn't depend on the mass
Then I used earth's gravitational field strength (g) 10 N/kg and the 2 m was the height provided in the question
you can try working and see if you get the same
Change the pressure and/or the temperature of the gas.
D. blue and white porcelain
Greatest number of major hurricanes making landfall in U.S.
Answer:
The baseball would seem to be traveling with a velocity 0.98918c of 98.92% of the speed of light as seen from the Earth.
Explanation:
This problem involves the relativistic effects. The pitcher throws the ball with a velocity u' which is 65% of the speed of light c i.e. u' = 0.65c
The velocity V of the spaceship in which he is practicing is 95% of the speed of light. Therefore V = 0.95c
So if the ball is thrown in the same direction as the direction of the spacecraft's motion, then classically, the velocities would add up and exceed the speed of light. But we know that this is not possible. It is prohibited by the special theory of relativity. So we need to consider the relativistic addition of velocities to calculate the velocity (u) of the ball as seen from the Earth.
The formula for relativistic addition of velocities is expressed below,

We know that u' = 0.65c and V = 0.95c. now we can calculate u which is the velocity of the baseball as seen from Earth as follows


u = 0.98918c
So the baseball would seem to be traveling with a velocity 0.98918c of 98.92% of the speed of light as seen from the Earth.