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.
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Answer:
vDP = 21.7454 m/s
θ = 200.3693°
Explanation:
Given
vDE = 7.5 m/s
vPE = 20.2 m/s
Required: vDP
Assume that
vDE to be in direction of - j
vPE to be in direction of i
According to relative motion concept the velocity vDP is given by
vDP = vDE - vPE (I)
Substitute in (I) to get that
vDP = - 7.5 j - 20.2 i
The magnitude of vDP is given by
vDP = √((- 7.5)²+(- 20.2)²) m/s = 21.7454 m/s
θ = Arctan (- 7.5/- 20.2) = 20.3693°
θ is in 3rd quadrant so add 180°
θ = 20.3693° + 180° = 200.3693°
A car has a mass of 900 kg and a truck has a mass of 1800 kg. In which of the following situations would they have the same momentum?A car has a mass of 900 kg and a truck has a mass of 1800 kg. In which of the following situations would they have the same momentum?
<h2>
Entire trip takes 1.22 seconds.</h2>
Explanation:
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Time, t = 0.866 s
Substituting
s = ut + 0.5 at²
s = 0 x 0.866 + 0.5 x 9.81 x 0.866²
s = 3.68 m
Halfway is 3.68 m
Total height = 2 x 3.68 = 7.36 m
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Time, t = ?
Displacement, s = 7.36 m
Substituting
s = ut + 0.5 at²
7.36 = 0 x t + 0.5 x 9.81 x t²
t = 1.22 s
Entire trip takes 1.22 seconds.
