The work done by the electric field in moving a charge is the negative of the potential energy difference between the two locations, which is the product between the magnitude of the charge q and the potential difference

:

The proton charge is

, and the two locations have potential of

and

, therefore the work is
The line is called equator.
Answer:
To create a second harmonic the rope must vibrate at the frequency of 3 Hz
Explanation:
First we find the fundamental frequency of the rope. The fundamental frequency is the frequency of the rope when it vibrates in only 1 loop. Therefore,
f₁ = v/2L
where,
v = speed of wave = 36 m/s
L = Length of rope = 12 m
f₁ = fundamental frequency
Therefore,
f₁ = (36 m/s)/2(12 m)
f₁ = 1.5 Hz
Now the frequency of nth harmonic is given in general, as:
fn = nf₁
where,
fn = frequency of nth harmonic
n = No. of Harmonic = 2
f₁ = fundamental frequency = 1.5 Hz
Therefore,
f₂ = (2)(1.5 Hz)
<u>f₂ = 3 Hz</u>
a). Perihelion . . . the point in Earth's orbit that's closest to the Sun.
We pass it every year early in January.
b). Aphelion . . . the point in Earth's orbit that's farthest from the Sun.
We pass it every year early in July.
c). Proxihelion . . . a made-up, meaningless word
d). Equinox . . . the points on the map of the stars where the Sun
appears to be on March 21 and September 21.
Explanation:
It is given that,
Mass of person, m = 70 kg
Radius of merry go round, r = 2.9 m
The moment of inertia, 
Initial angular velocity of the platform, 
Part A,
Let
is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

Here, 


Solving the above equation, we get the value as :

Part B,
The initial rotational kinetic energy is given by :



The final rotational kinetic energy is given by :



Hence, this is the required solution.