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3 years ago

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Physics

2 answers:

melamori03 [73]3 years ago

8 0

Answer:

the first one is neptune, second is saturn, thrid is uranus, and fourth is jupiter. just took the quiz ! hope this helps :)

Explanation:

Dmitriy789 [7]3 years ago

5 0

Answer:

the first one is: Neptune

second one: Saturn

third one: Uranus

fourth one: Jupiter

Explanation:

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A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b

Natasha_Volkova [10]

Answer:

Part a)

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part b)

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

Explanation:

Part a)

By Guass law we know that

$\int E. dA = \frac{q}{\epsilon_0}$

$E. 2\pi rL = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part b)

Outside the outer cylinder we will again use Guass law

$\int E. dA = \frac{q}{\epsilon_0}$

$E. 2\pi rL = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

4 0

3 years ago

Two narrow, parallel slits separated by 0.85 mm are illuminated by 600 nm light, and the viewing screen is 2.8 m away from the s

AURORKA [14]

Answer:

Phase difference = pi/4 radians

Explanation:

Given:

- The wavelength of incident light λ = 600 nm

- The split separation d = 0.85 mm

- Distance of screen from split plane L = 2.8 m

Find:

What is the phase difference between the two interfering waves on a screen, at a point 2.5 mm from the central bright fringe?

Solution:

- The phase difference can be evaluated by determining the type of interference that occurs at point y = 2.5 mm above central order. We will use the derived results from Young's double slit experiment.

sin ( Q ) = m*λ /d

m = d*sin(Q) / λ

- Where, m is the order number and angle Q is the angle for mth order of fringe from central bright fringe.

r = sqrt ( L^2 + 0.0025^ )

Where, r is the distance from split to the interference bright fringe.

r = sqrt(2.8^ + 0.0025^) = 2.8

sin(Q) = 0.0025 / 2.8

Hence. m = 0.00085*0.0025 / 2.8*(600*10^-9)

m = 1.26

- We know that constructive interference would occurred at m = 1 and destructive interference @ m = 1.5. They have a phase difference of pi/2 radians.

- The order number lies in between constructive and destructive interference i.e m ≈ 1.25 then the corresponding phase difference = 0.5*(pi/2).

Answer: Phase difference = pi/4 radians

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3 years ago

List the rock layers from oldest to youngest

MakcuM [25]

Answer:

I,J,h,g,f,e,d,c,b,a

Explanation:

8 0

3 years ago

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