Note:
The height of a high bar from the floor is h = 2.8 m (or 9.1 ft).
It is not provided in the question, so the standard height is assumed.
g = 9.8 m/s², acceleration due to gravity.
Note that the velocity and distance are measured as positive upward.
Therefore the floor is at a height of h = -2.8 m.
First dismount:
u = 4.0 m/s, initial upward velocity.
Let v = the velocity when the gymnast hits the floor.
Then
v² = u² - 2gh
v² = 16 - 2*9.8*(-2.8) = 70.88
v = 8.42 m/s
Second dismount:
u = -3.0 m/s
v² = (-3.0)² - 2*9.8*(-2.8) = 63.88 m/s
v = 7.99 m/s
The difference in landing velocities is 8.42 - 7.99 = 0.43 m/s.
Answer:
First dismount:
Acceleration = 9.8 m/s² downward
Landing velocity = 8.42 m/s downward
Second dismount:
Acceleration = 9.8 m/s² downward
Landing velocity = 7.99 m/s downward
The landing velocities differ by 0.43 m/s.
Answer:
flagpole
Explanation:
if it is about electricity then its flagpole
Given the distance r = 2/1000 m, the force between them F =
0.0104 N, the mass of the two object can be calculated using formula:
F = G(m1m2)/r^2 since the mass are equal F = G (m^2)/r^2
And where G = is the gravitational constant (6.67E-11 m3 s-2
kg-1)
The mass of the two objects are 24.96 kg
Answer:1. Roche limit
2.hydrogen
3.atmosphere
4.mercury
5.venus
6.when an object passes the Roche limit, the strength of gravity on the object increases. If the density of the planet is higher, then the object can break up farther away from the planet. If the density is lower, then the Roche limit is located closer to the planet
7.Farther our in the solar system, beyond the frost line, hydrogen was at a low enough temperature that it could condense. This allowed hydrogen to accumulate under gravity, eventually forming the Jovian planets
Explanation:
