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3 years ago

What is one use of the mineral gypsum fertilizer medicine wallboard electrical insulator

Physics

1 answer:

Semmy [17]3 years ago

7 0

Answer:

The main use for gypsum is in the production of Plaster of Paris and drywall products. Gypsum is also used as a soil conditioner.

Explanation:

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Light of wavelength 560 nm passes through a slit of width 0. 170 mm. (a) the width of the central maximum on a screen is 8. 00 m

faltersainse [42]

The distance between slit and the screen is 1.214m.

To find the answer, we have to know about the width of the central maximum.

<h3>How to find the distance between slit and the screen?</h3>

It is given that, wavelength 560 nm passes through a slit of width 0. 170 mm, and the width of the central maximum on a screen is 8. 00 mm.

We have the expression for slit width w as,

$w=\frac{2*wavelength*d}{a}$

where, d is the distance between slit and the screen, and a is the slit width.

Thus, distance between slit and the screen is,

$d=\frac{w*a}{2*wavelength} =\frac{8*10^{-3}*0.17*10^{-3}}{560*10^{-9}*2} \\\\d=1.214m$

Thus, we can conclude that, the distance between slit and the screen is 1.214m.

Learn more about the width of the central maximum here:

brainly.com/question/13088191

#SPJ4

3 0

2 years ago

A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit

Masja [62]

Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

Intrinsic impedance of media 1 = 377 ohms

Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c = 3 * 10⁸ Hz

Wavelength of the incident wave in air:

$\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m$

Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

$n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3$

$\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m$

b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

$n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

$n_1 = \sqrt{\frac{4\pi * 10^{-7} }{8.84 * 10^{-12} } }$

$n_1 = 377 \Omega$

The intrinsic impedance of media 2 is given as:

$n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

ϵr = 9

$n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }$

$n_2 = 125.68 \Omega$

c) The reflection coefficient,r and the transmission coefficient,t at the boundary.

Reflection coefficient, $r = \frac{n - n_{0} }{n + n_{0} }$

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

$r = \frac{3 - n_{0} }{3 + n_{0} }$

Transmission coefficient at the boundary, t = r -1

d) The amplitude of the incident electric field is $E_{0} = 10 V/m$

Maximum amplitudes in the total field is given by:

$E = tE_{0}$ and $E = r E_{0}$

E = 10r, E = 10t

3 0

3 years ago

Rami uses a disposable camera that has a flash. When he wants to take a picture, he holds a button and hears a rising whining so

ivann1987 [24]

<span>It stores energy and delivers it in a short burst.

The whirring sound is produced by the charging of the capacitor. A capacitor is an electrical component which is capable of storing charge. When the capacitor stores charge, it is storing energy. After doing so, the capacitor releases the electrical energy that it had stored as light energy, which is seen as the flash of the camera. It must do so in a burst, because the intensity of the flash is very high and would require a high amount of energy to maintain.
</span>

6 0

4 years ago

Two teams are playing tug of war. Team A pulls to the right with a force of 450 N. Team B pulls to the left with a force of 415

motikmotik

Answer:

35 N to the right.

Explanation:

450 is going to the right so you subtract what is going against it. Which gives you 35. And because 450 is bigger than 415, it'll be going to the right.

6 0

3 years ago

according to isaac newtons theory of gravitation the force exerted between two objects is dependent on- A- an objects weight. B-

puteri [66]

A an objects weight hope this helps! :D

3 0

3 years ago