Answer:
This type of star is called the "white dwarf." When a very massive star exhausts its nuclear fuel it explodes as a supernova.
Hope it helps
Answer:
The appropriate response will be "
". A further explanation is given below.
Explanation:
According to the question,
⇒
....(equation 1)
and,
⇒
....(equation 2)
Now,
On dividing both the equations, we get
⇒ 
⇒ 
⇒ 
Answer:
, the minus meaning west.
Explanation:
We know that linear momentum must be conserved, so it will be the same before (
) and after (
) the explosion. We will take the east direction as positive.
Before the explosion we have
.
After the explosion we have pieces 1 and 2, so
.
These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.
Since we know momentum must be conserved we have:

Which means (since we want
and
):

So for our values we have:

Answer:
<em>physical quantities that cannot be defined in terms of other quantities.</em>
<em>Examples</em><em>:</em><em> </em><em>Length - meter (m)Time - second (s)Amount of substance - mole (mole)Electric current - ampere (A)Temperature - kelvin (K)Luminous intensity - candela (cd)Mass - kilogram (kg)</em>
Answer:
2.82 s
Explanation:
The ball will be subject to the acceleration of gravity which can be considered constant. Therefore we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
Y0 is the starting position, 2.3 m in this case.
Vy0 is the starting speed, 13 m/s.
a will be the acceleration of gravity, -9.81 m/s^2, negative because it points down.
Y(t) = 2.3 + 13 * t - 1/2 * 9.81 * t^2
It will reach the ground when Y(t) = 0
0 = 2.3 + 13 * t - 1/2 * 9.81 * t^2
-4.9 * t^2 + 13 * t + 2.3 = 0
Solving this equation electronically gives two results:
t1 = 2.82 s
t2 = -0.17 s
We disregard the negative solution. The ball spends 2.82 seconds in the air.