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3 years ago

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Questlon 3 of 25

1 answer:

denpristay [2]3 years ago

5 0

CcccccccccccCccccccccccC

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OMG PLEASE HELPPPPPP

Alona [7]

Cccpkznsksmaajowneejkddp

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3 years ago

From the above diagram, which phase of the Moon will result in the greatest difference between high and low tide? A. New Moon B.

aliya0001 [1]

New moon, because the New Moon (A) because the sun and the moon work together most when it is there on the tides

8 0

3 years ago

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Three identical particles, q1, q2, and q3, each with charge q = 5.00 μC, are placed along a circle of radius r = 2.00 m at angle

wariber [46]

Answer:

11250 N/C

Direction: 0 deg counterclockwise from positive x-axis

Explanation:

$q$ = magnitude of charge on each particle = 5 μC = 5 x 10⁻⁶ C

$r$ = distance of each particle from center of circle = 2 m

$E$ = Magnitude of electric field at the center by each particle

Magnitude of electric field at the center by each particle is given as

$E = \frac{kq}{r^{2} }$

inserting the values

$E = \frac{(9\times10^{9} )(5\times10^{-6})}{2^{2} }\\E = 11.25\times10^{3} NC^{-1}$

From the diagram , we see that being equal and opposite, the electric fields due to charge q₁ and q₃ cancel out.

So net electric field at center is only due to charge q₂ direction towards positive x-direction

So

$E_{res}$ = Resultant electric field = 11250 N/C

Direction: 0 deg counterclockwise from positive x-axis

7 0

3 years ago

Light of wavelength 650 nm is normally incident on the rear of a grating. The first bright fringe (other than the central one) i

koban [17]

Answer:

A

$N = 1340.86 \ slits / cm$

B

$\theta = 15.7^o$

Explanation:

From the question we are told that

The wavelength is $\lambda = 650 \ nm = 650 *10^{-9} \ m$

The angle of first bright fringe is $\theta = 5^o$

The order of the fringe considered is n =1

Generally the condition for constructive interference is

$dsin (\theta ) = n * \lambda$

=> $d = \frac{1 * 650 *10^{-9 }}{ sin(5)}$

=> $d = 7.458 *10^{-6} \ m$

Converting to cm

$d = 7.458 *10^{-6} \ m = 7.458 *10^{-6} * 100 = 0.0007458 \ cm$

Generally the number of grating pre centimeter is mathematically represented as

$N = \frac{1}{d}$

=> $N = \frac{1}{0.0007458}$

=> $N = 1340.86 \ slits / cm$

Considering question B

From the question we are told that

The first wavelength is $\lambda_1 = 650 \ nm = 650 *10^{-9} \ m$

The second wavelength is $\lambda_2 = 429 \ m = 420 *10^{-9 } \ m$

The order of the fringe is $n = 2$

The grating is $N = 5000 \ slits / cm$

Generally the slit width is mathematically represented as

$d = \frac{1}{N }$

=> $d = \frac{1}{ 5000 }$

=> $d = 0.0002 \ c m = 2.0 *10^{-6} \ m$

Generally the condition for constructive interference for the first ray is mathematically represented as

$d sin(\theta_1) = n * \lambda_1$

=> $\theta_1 = sin^{-1} [\frac{ 2 * \lambda }{d}]$

=> $\theta_1 = sin^{-1} [\frac{ 2 * 650 *10^{-9} }{ 2*10^{-6}}]$

=> $\theta_1 = 40.5 ^o$

Generally the condition for constructive interference for the second ray is mathematically represented as

$d sin(\theta_2) = n * \lambda_2$

=> $\theta_2 = sin^{-1} [\frac{ 2 * \lambda_1 }{d}]$

=> $\theta_2 = sin^{-1} [\frac{ 2 * 420 *10^{-9} }{ 2*10^{-6}}]$

=> $\theta_2 = 24.8 ^o$

Generally the angular separation is mathematically represented as

$\theta = \theta_1 - \theta_1$

=> $\theta = 42.5^o - 24.8^o$

=> $\theta = 15.7^o$

4 0

3 years ago

What are the fundamental units involved in pascal<br>

Llana [10]

Answer:

<h2>FUNDAMENTAL UNITS INVOLVED ARE : NEWTON AND SECOND .</h2>

<h2>FORMULA OF PRESSURE = </h2>

<h2>P=F/A </h2>

4 0

4 years ago

Read 2 more answers