M = 30 g = 0.03 kg, the mass of the bullet
v = 500 m/s, the velocity of the bullet
By definition, the KE (kinetic energy) of the bullet is
KE = (1/2)*m*v²
= 0.5*(0.03 kg)*(500 m/s)² = 3750 J
Because the bullet comes to rest, the change in mechanical energy is 3750 J.
The work done by the wall to stop the bullet in 12 cm is
W = (1/2)*(F N)*(0.12 m) = 0.06F J
If energy losses in the form of heat or sound waves are ignored, then
W = KE.
That is,
0.06F = 3750
F = 62500 N = 62.5 kN
Answer:
(a) 3750 J
(b) 62.5 kN
Answer:
A. 200 J
Explanation:
The initial kinetic energy depends on the initial speed, while the gravitational potential energy depends on the height, both balls are thrown with the same initial speed and from the same height. Therefore, due to the law of conservation of energy, the balls must have the same mechanical energy (the sum of both energies) when both impact the ground. Since the potential energy is zero at this point, its final kinetic energy must also be the same.
The problem should only have one part to it, but this one has two.
Before I can do the mass/energy conversion, I have to go and
look up the proton mass for myself ... go out and collect the straw
to make my bricks, as it were. As if the fabulous bounty of 7 points
makes it worth it. They make us do everything around here.
OK. In my Physics book⁽¹⁾, the proton rest mass is
1.67 x 10⁻²⁷ kg.
The formula that relates mass to the equivalent energy is
E = m c² .
The method of applying the formula is known as "plug in what you know",
as follows:
E = (mass) x (speed of light)²
= (1.67 x 10⁻²⁷ kg) x (3 x 10⁸ m/s)²
= (1.67 X 10⁻²⁷ Kg) x (9 x 10¹⁶ m²/s²)
= (1.5 x 10⁻¹⁰) (kg-m²/s²)
= 1.5 x 10⁻¹⁰ joule .
____________________________________
⁽¹⁾ Halliday, David and Resnick, Robert, Physics , John Wiley & Sons,
Inc., 1960, inside front cover, "SELECTED PHYSICAL CONSTANTS".
the moon is illminated
tracing paper is not transparent
hope i helped
