Answer:
Work done.
Explanation:
The skater who lifts has to overcome the partner's weight. When lifted up by 1 meter, her potential energy increases by (mass)x(gravitational acceleration)x(1meter), which is the amount of work done.
(This all assumes lifting vertically and no other forces being part of the picture)
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:
The force becomes 16 times what it is now.
Explanation:
The formula for gravitational force is
F = G * m1 * m2 / r^2
When you do what you have described, you are setting a stage that not even the USS Enterprise (Star Trek) can get out of. The increase is huge.
If you double m1 and m2 and don't do anything to r, you've already increased the force by 4 times. (2m1 * 2m2 = 4 * m1 * m2)
But you are not finished. If you 1/2 the distance, you are again increasing the Force by 4 times. 1 / (2r) ^2 = 1/ 4* r^2
Because this is in the denominator, the 1/4 is going to flip to the numerator.
So the total increase is going to be 4 * (4 * m1 * m2) = 16 * m1 * m2.
Think about what that means. If you were out golfing, your drives would be roughly 1/16 times as far as they are now. Also you would be lugging around 16 times your weight around the golf course. My feeling is that you would never finish 5 holes at that rate.
Answer:
(a) q = 2.357 x 10⁻⁵ C
(b) Φ = 2.66 x 10⁶ N.m²/C
Explanation:
Given;
diameter of the sphere, d = 1.1 m
radius of the sphere, r = 1.1 / 2 = 0.55 m
surface charge density, σ = 6.2 µC/m²
(a) Net charge on the sphere
q = 4πr²σ
where;
4πr² is surface area of the sphere
q is the net charge on the sphere
σ is the surface charge density
q = 4π(0.55)²(6.2 x 10⁻⁶)
q = 2.357 x 10⁻⁵ C
(b) the total electric flux leaving the surface of the sphere
Φ = q / ε
where;
Φ is the total electric flux leaving the surface of the sphere
ε is the permittivity of free space
Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)
Φ = 2.66 x 10⁶ N.m²/C
