To solve this problem we will apply the theorem given in the conservation of energy, by which we have that it is conserved and that in terms of potential and kinetic energy, in their initial moment they must be equal to the final potential and kinetic energy. This is,
Replacing the 5100MJ for satellite as initial potential energy, 4200MJ for initial kinetic energy and 5700MJ for final potential energy we have that
Therefore the final kinetic energy is 3600MJ
Explanation:
The given data is as follows.
Velocity of bullet, = 814.8 m/s
Observer distance from marksman, d = 24.7 m
Let us assume that time necessary for report of rifle to reach the observer is t and will be calculated as follows.
t = (velocity in air = 343 m/s)
= 0.072 sec
Now, before the observer hears the report the distance traveled by the bullet is as follows.
=
= 58.66
= 59 (approx)
Thus, we can conclude that each bullet will travel a distance of 59 m.
Answer:
a=2.304×10¹⁶m/s²
Explanation:
Given data
Distance d=2.5 nm=2,5×10⁻⁹m
Mass of proton m=1.6×10⁻²⁷kg
charge of proton q=1.6×10⁻¹⁹C
To find
acceleration a
Solution
Apply the Coulombs Law
Where k is coulombs constant (k=9×10⁹Nm²/C²)
q=q₁=q₂
r=d
So