A solenoid has 450 loops each of radius 0.0254 m. The field increases from 0 T to 3.00 T in 1.55 s. What is the EMF generated in
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The distance between slit and the screen is 1.214m.
To find the answer, we have to know about the width of the central maximum.
<h3>How to find the distance between slit and the screen?</h3>- It is given that, wavelength 560 nm passes through a slit of width 0. 170 mm, and the width of the central maximum on a screen is 8. 00 mm.
- We have the expression for slit width w as,
where, d is the distance between slit and the screen, and a is the slit width.
- Thus, distance between slit and the screen is,
Thus, we can conclude that, the distance between slit and the screen is 1.214m.
Learn more about the width of the central maximum here:
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To solve this problem we will apply the linear motion kinematic equations. From the definition of the final velocity, as the sum between the initial velocity and the product between the acceleration (gravity) by time, we will find the final velocity. From the second law of kinematics, we will find the vertical position traveled.
Here,
v = Final velocity
= Initial velocity
g = Acceleration due to gravity
t = Time
At t = 4s, v = -30m/s (Downward)
Therefore the initial velocity will be
Now the position can be calculated as,
When it has the ground, y=0 and the time is t=4s,
Therefore the cliff was initially to 41.6m from the ground