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Ket [755]
3 years ago
8

The distance between two slits is 1.50 *10-5 m. A beam of coherent light of wavelength 600 nm illuminates these slits, and the d

istance between the slit and the screen is 2.00 m. What is the distance on the screen between the central bright fringe and the fourth-order bright fringe?
Physics
1 answer:
Fed [463]3 years ago
8 0

Answer: y = 2.4×10^-6m or y= 2.4μm

Explanation: The formulae for the distance between the central bright fringe to any other fringe in pattern is given as

y = R×mλ/d

Where y = distance between nth fringe and Central bright spot fringe.

m = position of fringe = 4

λ = wavelength of light= 600nm = 600×10^-9 m

d = distance between slits = 1.50×10^-5m

R = distance between slit and screen = 2m

y = 2 × 4 × 600×10^-9/2

y = 4800×10^-9/2

y = 2400 × 10^-9

y = 2.4×10^-6m or y= 2.4μm

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Joe is attempting to ramp his bicycle over a row of burning tires. If the ramp has a launch angle of 20 degrees, and Joe can ped
Luden [163]
Answer is 6 tires.

This is a projectile question.

First make sure units are consistent - express speed in m/s.

20 km/h = 20000m / 3600 s = 5.56 m/s

Assume the takeoff point of the ramp is at ground level (height, h, = 0m). We need to determine how long Joe is in the air, and use that time to calculate the horizontal distance he traveled.

Joe is traveling 5.56 m/s on a ramp angled at 20 degrees. There are vertical and horizontal components to his speed:

Vertical speed = 5.56sin20 = 1.90 m/s
Horizontal speed = 5.56cos20 = 5.22 m/s

An easy way to proceed is to calculate the time it takes for Joe’s vertical speed to reach 0m/s - this represents the time when Joe is at his maximum height and is therefore halfway through the trip. Double whatever time this is to find the total time of the trip. Remember he is decelerating due to gravity:

Time to peak:
a = Δv / Δt
-9.8 = -1.9 / Δt
Δt = 0.19s

Total trip time:
0.19 x 2 = 0.38s

Now that we have the total tome Joe is in the air, we can find the horizontal distance he traveled:

v = d / t
5.22 = d / 0.38
d = 1.98m

Now divide this total distance by the length of an individual tire to find the number of tires he will clear:

1.98 / 0.3 = 6.6 tires

Therefore he can jump 6 tires safely (he will land in the middle of the 7th tire).

Lots of steps I know but just try to think of the situation and keep track of the vertical and horizontal things!
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