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Luden [163]
3 years ago
5

A shift of one fringe in the michelson-morley experiment corresponds to a change in the round-trip travel time along one arm of

the interferometer by one period of vibration of light when the apparatus is rotated by 90 degrees. a particular interferometer has arm lengths of 8.50 meters, and is using light with a wavelength of 658 nm. what velocity through the ether would be deduced from a shift of one fringe?
Physics
1 answer:
8_murik_8 [283]3 years ago
5 0

Answer : [tex]v = 59.02\ km/s[/tex]

Explanation :  given that,

Arm length = 8.50 meters

Wavelength = 658 nm

Number of fringe = 1

We know that,

Formula of the speed is

v = c\sqrt\dfrac{N\lambda}{2L}

Now, put the values

v = 3\times10^{8}\ m/s\sqrt\dfrac{1\times658\times10^{-9}\ m}{2\times8.50\ m}

v = 590.21\times10^{2}\ m/s

v = 59.02\ km/s

Hence, this is the required solution.







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olya-2409 [2.1K]

Weight = (mass) x (acceleration of gravity)

Acceleration of gravity = 9.81 m/s² on Earth, 1.62 m/s² on the Moon.

The feather's weight is . . .

On Earth:  (0.0001 kg) x (9.81 m/s²) = <em>0.000981 Newton </em>

On the Moon:  (0.0001 kg) x (1.62 m/s²) = <em>0.000162 N</em>

The presence or absence of atmosphere makes no difference.  In fact, the numbers would be the same if the feather were sealed in a jar, or spinning wildly in a tornado, or hanging by a thread, or floating in a bowl of water or chicken soup.  Weight is just the force of gravity between the feather and the Earth.  It's not affected by what's around the feather, or what's happening to it.

6 0
3 years ago
Two identical bowling balls are rolling on a horizontal floor without slipping. The initial speed of both balls is v = 10 m/s. B
Lapatulllka [165]

Answer:

The difference between frictionless ramp and a regular ramp is that on a frictionless ramp the ball cannot roll it can only slide, but on a regular ramp the ball can roll without slipping.

We will use conversation of energy.

K_A_1 + U_A_1 = K_A_2 + U_A_2\\\frac{1}{2}I\omega^2 + \frac{1}{2}mv^2 + 0 = 0 + mgH_A

Note that initial potential energy is zero because the ball is on the bottom, and the final kinetic energy is zero because the ball reaches its maximum vertical distance and stops.

For the ball B;

K_B_1 + U_B_1 = K_B_2 + U_B_2

\frac{1}{2}I_B\omega^2 + \frac{1}{2}mv^2 + 0 = 0 + mgH_B

The initial velocities of the balls are equal. Their maximum climbing point will be proportional to their final potential energy. Since their initial kinetic energies are equal, their final potential energies must be equal as well.

Hence, both balls climb the same point.

Explanation:

4 0
3 years ago
A weightlifter lifts a 1250-N barbell 2 m in 3 s.How much power was used to lift the barbell?
Vlad [161]

The power is 833.3 W

Explanation:

First of all, we need to calculate the work done in lifting the barbell, which is equal to the change in gravitational potential energy of the barbell:

W=(mg)h

where

mg = 1250 N is the weight of the barbell

h = 2 m is the change in height

Substituting,

W=(1250)(2)=2500 J

Now we can calculate the power, which is equal to the work done per unit time:

P=\frac{W}{t}

where

W = 2500 J is the work done

t = 3 s is the time taken

Substituting,

P=\frac{2500}{3}=833.3 W

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

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30. The length of mercury thread when it is at 0°C, 100°C and at an unknown temperature 0 is 25mm, 225mm and 175mm respectively.
yulyashka [42]

Answer:

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Explanation:

175−25/225−25=x−0/100−0

150/200=x/100

x=150×100/200

= 75°C

3 0
3 years ago
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