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2 years ago

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In a laboratory experiment, a diffraction grating produces an interference pattern on a screen. If the number of slits in the gr

ating is increased, with everything else (including the slit spacing) the same, thenA.The fringes stay the same brightness and get closer together.B.The fringes stay the same brightness and get farther apart.C.The fringes stay in the same positions but get brighter and narrower.D.The fringes stay in the same positions but get dimmer and wider.E.The fringes get brighter, narrower, and closer together.

1 answer:

8_murik_8 [283]2 years ago

4 0

Answer:

C

Explanation:

Diffraction grating is an optics tool used when there is a need to separate light of different wavelengths using a high resolution. This tool is used in:

Optical pulse compressing devices
Spectrometers
Lasers

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An elevator does 9.75 x 10(4) J of work on a person riding up to another floor. How much energy does the person gain

Serjik [45]

Answer:

9.75 x 10^4 J

Explanation:

Work done, W = 9.75 x 10^4 J

According to the work energy theorem, the change in kinetic energy is equal to the work done by all the forces.

So, here work done is 9.75 x 10^4 J so the change in kinetic energy is 9.75 x 10^4 J.

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2 years ago

Suppose you have a total charge qtot that you can split in any manner. Once split, the separation distance is fixed. How do you

dybincka [34]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know from the Coulomb's Law that, Coulomb's force is directly proportional to the product of two charges q1 and q2 and inversely proportional to the square of the radius between them.

So,

F = $\frac{Kq1q2}{r^{2} }$

Now, we are asked to get the greatest force. So, in order to do that, product of the charges must be greatest because the force and product of charges are directly proportional.

Let's suppose, q1 = q

So,

if q1 = q

then

q2 = Q-q

Product of Charges = q1 x q2

Now, it is:

Product of Charges = q x (Q-q)

So,

Product of Charges = qQ - $q^{2}$

And the expression qQ - $q^{2}$ is clearly a quadratic expression. And clearly its roots are 0 and Q.

So, the highest value of the quadratic equation will be surely at mid-point between the two roots 0 and Q.

So, the midpoint is:

q = $\frac{Q + 0}{2}$

q = Q/2 and it is the highest value of each charge in order to get the greatest force.

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3 years ago

You've recently read about a chemical laser that generates a 20.0-cm-diameter, 26.0 MW laser beam. One day, after physics class,

aksik [14]

Answer :

(a). The speed of the block is 0.395 m/s.

(b). No

Explanation :

Given that,

Diameter = 20.0 cm

Power = 26.0 MW

Mass = 110 kg

diameter = 20.0 cm

Distance = 100 m

We need to calculate the pressure due to laser

Using formula of pressure

$P_{r}=\dfrac{I}{c}$

$P_{r}=\dfrac{P}{Ac}Put the value into the formula[tex]P_{r}=\dfrac{26.0\times10^{6}}{\pi\times(10\times10^{-2})^2\times3\times10^{8}}$

$P_{r}=2.75\ N/m^2$

We need to calculate the force

Using formula of force

$F=P\times A$

$F=P\times \pi r^2$

Put the value into the formula

$F=2.75\times\pi (0.01)^2$

$F=0.086\ N$

We need to calculate the acceleration

Using formula of force

$F=ma$

Put the value into the formula

$0.086=110\times a$

$a=\dfrac{0.086}{110}$

$a=0.000781\ m/s^2$

$a=7.81\times10^{-4}\ m/s^2$

(a). We need to calculate speed of the block

Using equation of motion

$v^2=u^2+2ad$

Put the value into the formula

$v=\sqrt{2\times7.81\times10^{-4}\times100}$

$v=0.395\ m/s$

(b). No because the velocity is very less.

Hence, (a). The speed of the block is 0.395 m/s.

(b). No

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2 years ago

I NEED HELP!!! PLS HELP ME MARK U BRAINLIEST!!

FinnZ [79.3K]

Answer:

The answer is X

Explanation:

Cause the highest points will most likely have the most potential energy

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2 years ago

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0

2 years ago