Answer:
Explanation:
Potential energy at the surface
= -GMm / R
= -6.6 x 10⁻¹¹ x 6 x 10²³ m / 3400 x 10³
= -1.1647 x 10⁷ m
If v be required velocity
Total energy at surface
= 1/2 m v² -1.1647 x 10⁷ m
= energy at farr off = 1/2 m x 3000²
1/2 m v² - -1.1647 x 10⁷ m = 1/2 m x 9 x 10⁶
1/2 v²= 1.1647 x 10⁷ + .45 x 10⁷
= 1.6147 x 10⁷
v² = 2.3303 x 10⁷
v = 4.827 x 10³ m / s
b ) For final speed zero at infinity
1/2 m v² - -1.1647 x 10⁷ m = 0
1/2 v² = 1.1647 x 10⁷
v² = 2.3294 x 10⁷
v = 4.826 x 10³ m /s
At theheight where it starts, just before it's dropped, the ball has
some potential energy. The higher that spot is, the more potential
energy the ball has. After the drop, whenever the ball is lower than
the height from which it was dropped, it has less potential energy, and
the missing potential energy shows up as kinetic energy ... motion.
This is the whole idea of the roller coaster. A machine drags it up to
the top of the first hill, giving it lots of potential energy. After that, as
long as it doesn't try to rise higher than the first hill, it never runs out
of energy, and keeps going.
A). and B).
The ball keeps going forward until it rises again to the same height it
was dropped from ... on the other side. Then it stops and falls back.
C). The ball can never rise higher than the height it was dropped from.
If the hump in the middle is the same height as the drop-height, then
the ball stops right there, and falls back.
D). Same as B). As long as the track inside the loop is never higher
than the droop-height, the ball just keeps going forward.
E). Same idea. Here it looks like the drop-height is the same as the
top of the loop. The ball can't rise higher than it was dropped from,
so it gets as far as the top of the loop and stops there. From there,
I think it drops straight down from the top of the loop, instead of
following the curve.
Density is mass divided by volume. rho=m/v. So, v=m/rho. In frank's case this is 80/8 = 10 cm^3.
Answer:
Velocity is the displacement (change In position of an an object it is similar to distance ) divided by the time taken. SI unit m/s (metre per second)
The number of charge drifts are 3.35 X 10⁻⁷C
<u>Explanation:</u>
Given:
Potential difference, V = 3 nV = 3 X 10⁻⁹m
Length of wire, L = 2 cm = 0.02 m
Radius of the wire, r = 2 mm = 2 X 10⁻³m
Cross section, 3 ms
charge drifts, q = ?
We know,
the charge drifts through the copper wire is given by
q = iΔt
where Δt = 3 X 10⁻³s
and i = 
where R is the resistance
R = 
ρ is the resistivity of the copper wire = 1.69 X 10⁻⁸Ωm
So, i = 
q = 
Substituting the values,
q = 3.14 X (0.02)² X 3 X 10⁻⁹ X 3 X 10⁻³ / 1.69 X 10⁻⁸ X 0.02
q = 3.35 X 10⁻⁷C
Therefore, the number of charge drifts are 3.35 X 10⁻⁷C
