Answer:
The distance between the two slits is 40.11 μm.
Explanation:
Given that,
Frequency 
Distance of the screen l = 88.0 cm
Position of the third order y =3.10 cm
We need to calculate the wavelength
Using formula of wavelength

where, c = speed of light
f = frequency
Put the value into the formula


We need to calculate the distance between the two slits


Where, m = number of fringe
d = distance between the two slits
Here, 
Put the value into the formula



Hence, The distance between the two slits is 40.11 μm.
Answer:
Frequency required will be 2421.127 kHz
Explanation:
We have given inductance 
Current in the inductor 
Voltage v = 13 volt
Inductive reactance of the circuit 

We know that


f = 2421.127 kHz
Answer:
They experience the same magnitude impulse
Explanation:
We have a ping-pong ball colliding with a stationary bowling ball. According to the law of conservation of momentum, we have that the total momentum before and after the collision must be conserved:
where is the initial momentum of the ping-poll ball
is the initial momentum of the bowling ball (which is zero, since the ball is stationary)
is the final momentum of the ping-poll ball
is the final momentum of the bowling ball
We can re-arrange the equation as follows or
which means (1) so the magnitude of the change in momentum of the ping-pong ball is equal to the magnitude of the change in momentum of the bowling ball.
However, we also know that the magnitude of the impulse on an object is equal to the change of momentum of the object:
(2) therefore, (1)+(2) tells us that the ping-pong ball and the bowling ball experiences the same magnitude impulse: