Power is joules per second or J/s when work is measured in joules and time in seconds. The basic unit of power, 1 J/s is called a watt (W), named after James Watt who made important improvements to the steam engine. By definition, a watt is the consumption of one joule of energy per second.
Answer: Ok, first lest see out problem.
It says it's a Long cylindrical charge distribution, So you can ignore the border effects on the ends of the cylinder.
Also by the gauss law we know that E¨*2*pi*r*L = Q/ε0
where Q is the total charge inside our gaussian surface, that will be a cylinder of radius r and heaight L.
So Q= rho*volume= pi*r*r*L*rho
so replacing : E = (1/2)*r*rho/ε0
you may ask, ¿why dont use R on the solution?
since you are calculating the field inside the cylinder, and the charge density is uniform inside of it, you don't see the charge that is outside, and in your calculation actuali doesn't matter how much charge is outside your gaussian surface, so R does not have an effect on the calculation.
R would matter if in the problem they give you the total charge of the cylinder, so when you only have the charge of a smaller r radius cylinder, you will have a relation between r and R that describes how much charge density you are enclosing.
Answer:
Explanation:
Given:
Initial θ = 0 rad (from rest)
Final θ = 14.3 rad
Time, t = 5 s
B.
Angular velocity, w = dθ / dt
= (14.3 - 0)/5
= 2.86 rad/s
A.
Acceleration, ao = dw/dt
Initial angular velocity, wi = 0 rad/s (from rest)
Final angular velocity, wf = 2.86 rad/s
a = (2.86 - 0)/5
= 0.572 rad/s^2
The correct answer is 432, and 720.
The thickness of a film is t= 360nm
the refractive index of oil n₀t = (m +1/2) λ
For m =0
λ = 4n₀t
= 4(1.50)(360)
= 2160nm
for m = 1
λ = 4n₀t
= 4(1.50)(360)/3
= 720nm
m = 2
λ = 4n₀t/5 = 4(1.50)(360)/5
= 432nm
The wavelength which are most strongly reflected are
432nm, 720nm.
Answer:
25 m/s in the opposite direction with the ship recoil velocity.
Explanation:
Assume the ship recoil velocity and velocity of the cannon ball aligns. By the law of momentum conservation, the momentum is conserved before and after the shooting. Before the shooting, the total momentum is 0 due to system is at rest. Therefore, the total momentum after the shooting must also be 0:
where 
are masses of the ship and ball respectively. 
are the velocities of the ship and ball respectively, after the shooting.
So the cannon ball has a velocity of 25 m/s in the opposite direction with the ship recoil velocity.
