To solve this problem we will apply the concepts related to load balancing. We will begin by defining what charges are acting inside and which charges are placed outside.
PART A)
The charge of the conducting shell is distributed only on its external surface. The point charge induces a negative charge on the inner surface of the conducting shell:
. This is the total charge on the inner surface of the conducting shell.
PART B)
The positive charge (of the same value) on the external surface of the conducting shell is:
The driver's net load is distributed through its outer surface. When inducing the new load, the total external load will be given by,
Given gravitational potential energy when he's lifted is 2058 J.
Kinetic energy is transferred to the person.
Amount of kinetic energy the person has is -2058 J
velocity of person = 7.67 m/s².
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Explanation:</h3>
Given:
Weight of person = 70 kg
Lifted height = 3 m
1. Gravitational potential energy of a lifted person is equal to the work done.
Gravitational potential energy is equal to 2058 Joules.
2. The Gravitational potential energy is converted into kinetic energy. Kinetic energy is being transferred to the person.
3. Kinetic energy gained = Potential energy lost = 
Kinetic energy gained by the person = (-2058 kg.m/s²)
4. Velocity = ?
Kinetic energy magnitude= 
Solving for v, we get
The person will be going at a speed of 7.67 m/s².
Answer:
It will be cut in half
Explanation:
The diffraction of a slit is given by the formula
a sin θ = m where
a = width of the slit,
λ = wavelength and
m = integer that determines the order of diffraction.
Next we divide both sides by a, we have
sin θ = m λ / a
Also, recall that
a’ = 2 a
Then we substitute in the previous equation
2asin θ' = m λ, if divide by 2a, we have
sin θ' = (m λ / 2a).
Now again, from the first equation, we said that sin θ = m λ / a, so we substitute
sin θ ’= sin θ / 2
Then we use trigonometry to find the width, we say
tan θ = y / L
Since the angle is small, we then have
tan θ = sin θ / cos θ
tan θ = sin θ, this then means that
sin θ = y / L
we will then substitute
y’ / L = y/L 1/2
y' = y / 2
this means that when the slit width is doubled the pattern width will then be halved
Answer:
The answer to the questions is;
In terms of standing waves, the listener moves from a location with high amplitude to one with lower amplitude or vibration (anti-node to node)
The distance 4.1 cm is equivalent to λ/4
Explanation:
For standing waves we have is a stationary wave comprising of two opposite direction moving waves that have equal amplitude and frequency, resulting in the superimposition of the waves. As such certain points are fixed along the wave path that is the peaks amplitude of the wave oscillation is constant at a particular point. A node occurring at a point and an anti-node occurring at another fixed point
When the listener moves 4.1 cm he or she has left the anti-node to the node hence the faintness of the sound
The distance from the node to the anti-node is 1/4 wavelength, or 1/4×λ
Therefore 4.1 cm is λ/4
