Answer:
Explanation:
Let the required velocity of rocket be v .
We shall use the formula of time dilation to find the velocity of rocket .
t =
t = 430
t' = 38


= .0078
=.9922
= .996
v = .996 x 3 x 10⁸ m /s
= 2.988 x 10⁸ m /s
B )
Kinetic energy of rocket
= 1/2 m v²
= .5 x 20000 x (2.988 x 10⁸ )²
= 8.9 x 10²⁰ J .
C )
This energy is 8.9 times the energy requirement of United states in the year 2005 .
The moon would be bright and the earth would be darker because the sun is on the opposite side of the earth at that time and the light from it is reflecting off the moon to produce light upon the nigh also.......
You wouldn’t see the sun a night...
Unless you lived in the north/south pole
Given:
object = 20kg
terminal speed of object = 80 m/s
According to the problem, drag force is proportional to speed, so Fd = kv ; k is some constant
At terminal velocity Vt: Fg = Fdmg = kVtk = mg / Vt = (20.0)(9.8)/(80.0) = 2.45 kg/s
<span>Fd = kv = 2.45v</span>
Fd = 2.45 (30.0) = 73.5 N
They all eventually burn out or explode.
Answer:
Star A would be brighter than Star B
Explanation:
The apparent brightness of a star as perceived on Earth is dependent on its temperature, size, luminosity and distance from the Earth. Apparent brightness is the visible brightness to the eye at the surface of the Earth, while luminosity is the true brightness at the surface of the star.
A hotter star will radiate more energy per second per meter square of surface area. A larger star will have a greater surface area for radiation of energy, thus increasing the luminosity. For two identical stars, the difference in apparent brightness will be dependent on their distances from Earth.
Brightness and distance from earth have an inverse square relationship.
∝
Assuming the star is a point source of radiation, as distance from the source is increased, the radiation is distributed over a surface proportional to the distance form the source. As distance is further increased, the radiation is distributed over a larger surface area reducing the effective luminosity.
If one star (Star B) is twice as far from the earth as the first (Star A), the brightness of Star B will be
of Star A.
Thus, Star B will appear to be a quarter of the brightness of Star A. Or, Star A will appear to be 4 times as bright as Star B.