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spayn [35]
2 years ago
7

A piece of fine fiber with a diameter of =6.5 m is used to prop apart the edges of two perfectly flat 3.3-cm-long pieces of glas

s (see diagram). When the setup is illuminated from above with light of wavelength =590 nm , an interference pattern of alternating bright and dark bands will be seen in the reflected light. If the setup is viewed from high above, how many dark bands will be seen?
Physics
1 answer:
Anna71 [15]2 years ago
3 0

We need frequency:-

\\ \rm\Rrightarrow \nu=\dfrac{c}{\lambda}

\\ \rm\Rrightarrow \nu=\dfrac{3\times 10^8m/s}{590\times 10^{-9}m}

\\ \rm\Rrightarrow \nu=0.0051\times 10^{17}Hz

\\ \rm\Rrightarrow \nu=5.1\times 10^{14}Hz

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On a hot summer day a young girl swings on a rope above the local swimming hole. when she lets go of the rope her initial veloci
weqwewe [10]
Refer to the diagram shown below.
 h =  height of the girl above water when she lets go of the rope.

The launch velocity is 22.5 m/s at 35° to the horizontal. Therefore the vertical component of the velocity is
v = 22.5 sin(35°) = 12.9055 m/s.

The time of flight is t = 1.10 s before the girl hits the surface of the water at a height of  -h.
Therefore
-h = (12.9055 m/s)*(1.10 s) - (1/2)*(9.8 m/s²)*(1.10 s)²
-h = 8.267 m
    = 8.3 m (nearest tenth)

Answer: 
When the girl let go of the rope, she was about 8.3 m  above the surface of the water.

5 0
4 years ago
While visiting the planet Mars, Moe leaps straight up off the surface of the Martian ground with an initial velocity of 105 m/s
yawa3891 [41]

Answer:

L = 0 m

Therefore, the cricket was 0m off the ground when it became Moe’s lunch.

Explanation:

Let L represent Moe's height during the leap.

Moe's velocity v at any point in time during the leap is;

v = dL/dt = u - gt .......1

Where;

u = it's initial speed

g = acceleration due to gravity on Mars

t = time

The determine how far the cricket was off the ground when it became Moe’s lunch.

We need to integrate equation 1 with respect to t

L = ∫dL/dt = ∫( u - gt)

L = ut - 0.5gt^2 + L₀

Where;

L₀ = Moe's initial height = 0

u = 105m/s

t = 56 s

g = 3.75 m/s^2

Substituting the values, we have;

L = (105×56) -(0.5×3.75×56^2) + 0

L = 0 m

Therefore, the cricket was 0m off the ground when it became Moe’s lunch.

5 0
3 years ago
A timeline is necessary a. For every goal c. For every life goal b. For every academic goal d. For every high school goal please
kiruha [24]
I’d say for every life goal
5 0
3 years ago
Read 2 more answers
Which can cause blockage of the ear canal, leading to hearing loss? aging disease earwax noise
Ivahew [28]

Explanation:

Earwax helps to protect your ears from infection so if you like chewing gum, this helps in removing the wax. But this might not happen at your young age. So if you grow and still likes chewing gum it will remove the wax completely thereby leading to loss on hearing

5 0
3 years ago
Dolphins emit clicks of sound for communication and echolocation. A marine biologist is monitoring a dolphin swimming in seawate
LenKa [72]

Answer:

12.64968 Hz

Explanation:

v = Velocity of sound in seawater = 1522 m/s

u = Velocity of dolphin = 7.2 m/s

f' = Actual frequency = 2674 Hz

From Doppler effect we get the relation

f=f'\frac{v-u}{v}\\\Rightarrow f=2674\frac{1522-7.2}{1522}\\\Rightarrow f=2661.35032\ Hz

The frequency that will be received is 2661.35032 Hz

The difference in the frequency will be

2674-2661.35032=12.64968\ Hz

6 0
4 years ago
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