Answer:
Yes
Explanation:
Eclipses: Eclipses are also known as game of shadows where one object comes between the star(light source) and another object in a straight line such that the shadow of one object falls on other object. This can occur when the apparent size of the star and the object is almost same.
Talking about the Earth, the geometry is such that the Moon and the Sun are of same apparent size as seen from the Earth. Thus Lunar and Solar eclipse can be seen from the Earth. If we were to go on any other planet the same phenomenon can be seen provided the apparent size of moon and the Sun from that planet is same.
We have seen and recorded many such eclipses on Jupiter. These are from the perspective of Earth. When the moons of Jupiter comes exactly between the Sun and Jupiter the shadow of moon will fall on Jupiter. The places where the shadow falls, one will see a solar eclipse.
Answer:11 km/s
Explanation:
Given
Escape velocity at the surface of earth is 11 km/s
Escape velocity is given by

Escape velocity at the surface of earth
--------------------1
If Escape velocity is three times and the radius is also the three times


i.e. 
Answer:
No
Explanation:
There is no limit to how fast the universe can expand, says physicist Charles Bennett of Johns Hopkins University. Einstein's theory that nothing can travel faster than the speed of light in a vacuum still holds true, because space itself is stretching, and space is nothing.
Answer:
Temperature after ignition=7883.205 K
Explanation:
The number of moles is,
n=PV/RT
=(1.18x10^6)(47.9x10^-6)/8.314(325)
= 0.0209 moles
a) In this process volume is constant
Q=U
=nCv.dT
dT= Q/nCv
=1970/(1.5x8.314)(0.0209)
= 7558.205 K
The final temperature is,
= 7558.205+325
= 7883.205 K