The first law, which deals with changes in the internal energy, thus becomes 0 = Q - W, so Q = W.
If the system does work, the energy comes from heat flowing into the system from the reservoir; if work is done on the system, heat flows out of the system to the reservoir
Answer:
If the particle is an electron
If the particle is a proton,
Explanation:
Initial speed at the origin,
to +ve x-axis
The particle crosses the x-axis at , x = 1.5 cm = 0.015 m
The particle can either be an electron or a proton:
Mass of an electron,
Mass of a proton,
The electric field intensity along the positive y axis , can be given by the formula:
If the particle is an electron:
If the particle is a proton:
Answer:
a)
<u>get mass of planet:</u>
ρ = M / V
V = 4/3 * R_1^3
M = ρ * V
M = ρ * 4/3 * R_1^3
<u>equate force equations:</u>
F = (GMm) / r^2 // r = R_2
F = ma
a = v^2/R_2
F = m * (v^2/R_2)
m * (v^2/R_2) = (GMm) / R_2^2
<u>plug in and solve v^2:</u>
m * (v^2/R_2) = (G * (ρ * 4/3 * R_1^3) *m) / R_2^2
v^2 = (G * ρ * (4/3) * π * R_1^3) / R_2
<u>put into kinetic energy equation:</u>
K = 1/2 * m * v^2
K = 1/2 * m * (G * ρ * (4/3) * π * R_1^3) / R_2
B)
<u>givens:</u>
U = -(GmM) / R_2
<u>plug in mass of planet:</u>
U = -(G * m * ρ * 4/3 * R_1^3) / R_2
C)
<u>use previous equations and do some algebra:</u>
K/U = (1/2 * m * (G * ρ * (4/3) * π * R_1^3) / R_2) * -(R_2 / (G * m * ρ * 4/3 * R_1^3))
K/U = -1/2
In this graph, what is the displacement of the particle in the last two seconds?of the particle in the last two seconds?
<span>0.2 meters
2 meters
4 meters
6 meters</span>
In this graph, the displacement of the particle in the last two seconds is 2 meters.