Answer:
13.7m
Explanation:
Since there's no external force acting on the astronaut or the satellite, the momentum must be conserved before and after the push. Since both are at rest before, momentum is 0.
After the push
Where 
is the mass of the astronaut, 
is the mass of the satellite, 
is the speed of the satellite. We can calculate the speed 
of the astronaut:
So the astronaut has a opposite direction with the satellite motion, which is further away from the shuttle. Since it takes 7.5 s for the astronaut to make contact with the shuttle, the distance would be
d = vt = 1.83 * 7.5 = 13.7 m
Answer:
a).11.546J
b).2.957kW
Explanation:
Using Inertia and tangential velocity
a).
Now using Inertia an w
average power=
b).
power=t*w
P=
P=2.957 kW
Answer
Ceres, Pluto, and Eris are classified as DWARF PLANET.
A) Leftover planetesimals inside the frost line are known as ASTEROIDS.
B) METEORITES are the pieces of Asteroids which are fallen on the earth's surface.
C) COMETS are the objects which are visible with long tails.
D) COMETS are also the leftover planetesimals that are occupied by the jovian planets and are formed in the solar system.
E) Meteor showers are associated with debris from COMETS
Answer:
The electric flux is 
Explanation:
Given:
- Radius of the disc R=0.50 m
- Angle made by disk with the horizontal
- Magnitude of the electric Field
The flux of the Electric Field E due to the are dA in space can be found out by using Gauss Law which is as follows
where
is the total Electric Flux- E is the Electric Field
- dA is the Area through which the electric flux is to be calculated.
Now according to question we have
Hence the electric flux is calculated.
Clever problem.
We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks. So if Fork-A is 256 Hz and the beat is 6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz. But which one is it ?
Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz. That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.
If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.
The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz. While it was loaded with wax, it was 261 Hz.
