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Luden [163]
2 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]2 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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svet-max [94.6K]
We have: F = m×a
Here, m = 90 Kg
a = 15 m/s²

Substitute their values into the expression:
F = 90 × 15
F = 1350 N

In short, Your Answer would be Option D

Hope this helps!
8 0
3 years ago
The amount of energy transported by a wave is most closely related to the wavelength of the wave. True or False?
Artemon [7]
I think its false............ Im pretty sure
3 0
2 years ago
Read 2 more answers
1. Determine the image distance in each of the following.
miskamm [114]

It is given that for the convex lens,

Case 1.

u=−40cm

f=+15cm

Using lens formula

v

1

−

u

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=

f

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v

1

−

40

1

=

15

1

v

1

=

15

1

−

40

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v=+24.3cm

The image in formed in this case at a distance of 24.3cm in left of lens.

Case 2.

A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens i.e. the source is placed at the focus of mirror, then the rays after reflection becomes parallel for the lens such that

u=∞

f=15cm

Now, using mirror’s formula

v

1

+

u

1

=

f

1

v

1

+

∞

1

=

15

1

v=+15cm

The image is formed at a distance of 15cm in left of mirror

6 0
3 years ago
A bicycle rider pushes a 13kg bicycle up a steep hill. the incline is 24 degree and the road is 275m long. the rider pushes the
Digiron [165]

Answer:

A. W = 6875.0 J.

B. W = -14264.6 J.

Explanation:

A. The work done by the rider can be calculated by using the following equation:

W_{r} = |F_{r}|*|d|*cos(\theta_{1})

Where:                

F_{r}: is the force done by the rider = 25 N

d: is the distance = 275 m

θ: is the angle between the applied force and the distance

Since the applied force is in the same direction of the motion, the angle is zero.

W_{r} = |F_{r}|*|d|*cos(0) = 25 N*275 m = 6875.0 J

Hence, the rider does a work of 6875.0 J on the bike.

B. The work done by the force of gravity on the bike is the following:

W_{g} = |F_{g}|*|d|*cos(\theta_{2})  

The force of gravity is given by the weight of the bike.

F_{g} = -mgsin(24)     

And the angle between the force of gravity and the direction of motion is 180°.

W_{g} = |mgsin(24)|*|d|*cos(\theta_{2})  

W_{g} = 13 kg*9.81 m/s^{2}*sin(24)*275 m*cos(180) = -14264.6 J  

The minus sign is because the force of gravity is in the opposite direction to the motion direction.

Therefore, the magnitude of the work done by the force of gravity on the bike is 14264.6 J.  

I hope it helps you!                                                                                          

3 0
2 years ago
A uniform, solid, 2000.0 kgkg sphere has a radius of 5.00 mm. Find the gravitational force this sphere exerts on a 2.10 kgkg poi
mars1129 [50]

Answer:

0.0110284391534\ N

0.0653784219002\ N

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

m_1 = Mass of sphere = 2000 kg

m_2 = Mass of other sphere = 2.1 kg

r = Distance between spheres

Force of gravity is given by

F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 2000\times 2.1}{(5.04\times 10^{-3})^2}\\\Rightarrow F=0.0110284391534\ N

The gravitational force is 0.0110284391534\ N

F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 2000\times 2.1}{(2.07\times 10^{-3})^2}\\\Rightarrow F=0.0653784219002\ N

The gravitational force is 0.0653784219002\ N

6 0
3 years ago
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