what is amplitudes relationship between sound and amplitudes ​

Answers

Answer 1

Answer:

A sound wave's amplitude relates to changes in pressure. The sound is perceived as louder if the amplitude increases, and softer if the amplitude decreases. This is illustrated below. DOSITS short video on amplitude.


Related Questions

A baseball with a mass of 142 g is thrown horizontally with a speed of 38.6 m/s (86 mi/h) at a bat. The ball is in contact with the bat for 1.00 ms and then travels straight back at a speed of 46.1 m/s (103 mi/h). Determine the average force (in N) exerted on the ball by the bat. Neglect the weight of the ball (it is much smaller than the force of the bat) and choose the direction of the incoming ball to be positive. (Indicate the direction with the sign of your answer.)

Answers

Answer:

-11.600N

Explanation:

lol

(will give brainliest to whoever is correct and shows reasoning) What is the acceleration of an object that has a velocity of 60m/s and is moving in a circle of radius 50m?

Answers

Answer:

a = 72 m/s^2

Explanation:

Since the object is moving in circle, the acceleration points TOWARDS the centre of the circle (it is centripetal acceleration)

a = v^2 / r

a = 60^2 / 50

a = 72 m/s^2

A 40 kg rock is rolling toward a town at 4 m/s after an earthquake. Calculate the KE.

Be sure to show your work and include units!

Use the formula KE = 1/2mv2 will brainlist

Answers

Answer:

320 J

Explanation:

From the question,

KE = 1/2mv².................. Equation 1

Where KE = Kinetic Energy, m = mass of the rock, v = velocity of the rock

Given: m = 40 kg, v = 4m/s

Substitute these values into equation 1

KE = 1/2(40)(4²)

KE = 20×16

KE = 320 J

Hence the kinetic energy of the rock is 320 J

Four children climb on a carousel that is initially at rest. If the carousel accelerates to 0.4 rad/s within 10 seconds, what is the angular acceleration? If the carousel has a radius of 20 meters, what is the linear velocity of the children?

Answers

Answer:

If the angular acceleration is a constant A, then this constant times the time will be equal to the angular velocity.

We know that after accelerating for 10 seconds, the angular velocity is 0.4 rad/s

Then:

A*10s = 0.4 rad/s

If we solve this for A we get:

A = (0.4 rad/s)/10s = 0.04 rad/s^2

This is the angular acceleartion.

Now we assume that the angular speed remains constant at 0.4 rad/s, and we know that the radius of the carousel is 20m, we want to find the linear velocity of the children.

We know that for an angular velocity W and for a radius R, the linear velocity is:

V = R*W

Then in this case the linear velocity is:

V = 20m*(0.4 s^(-1)) = 8 m/s

During which phase is the moon not visible?
A) Full Moon
B) First quarter
C) New moon
D) Waxing crescent

Answers

Answer:

they are right it is a new moon

Explanation:

took the test

You can hear sounds transmitted through alr, water, or steel but not through the empty vacuum of space. What type of wave carries sound?

gravity

light

mechanical

electromagnetic

Answers

Electromagnetic because Light is an electromagnetic wave and can travel through the vacuum of outer space.

How much energy would be required to move the earth into a circular orbit with a radius 2.0 kmkm larger than its current radius

Answers

Answer:

[tex]3.52\times 10^{25}\ \text{J}[/tex]

Explanation:

G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]

M = Mass of Sun = [tex]1.989\times 10^{30}\ \text{kg}[/tex]

m = Mass of Earth = [tex]5.972\times 10^{24}\ \text{kg}[/tex]

[tex]r_i[/tex] = Initial radius of orbit = [tex]1.5\times 10^{11}\ \text{m}[/tex]

[tex]r_f[/tex] = Final radius of orbit = [tex]((1.5\times 10^{11})+2\times 10^3)\ \text{m}[/tex]

Energy required is given by

[tex]E=\dfrac{1}{2}\Delta U\\\Rightarrow E=\dfrac{GMm}{2}(\dfrac{1}{r_i}-\dfrac{1}{r_f})\\\Rightarrow E=\dfrac{6.674\times 10^{-11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{2}(\dfrac{1}{1.5\times 10^{11}}-\dfrac{1}{(1.5\times 10^{11})+2\times 10^3})\\\Rightarrow E=3.52\times 10^{25}\ \text{J}[/tex]

The energy required would be [tex]3.52\times 10^{25}\ \text{J}[/tex].

३.रात में घूमने वाला write one word substitute ​

Answers

Explanation:

रात में घूमने वाला arthaarat निशाचर

DUE TODAY BEST ANSWER GET BRAINLIE
Which statements explain the relationship between mass, velocity and kinetic energy? * Select the TWO (2) that apply.*

Consider the equation to calculate kinetic energy: KE = ½mv2


Kinetic energy increases if either the mass or the velocity of the object increases or if both increase.

Kinetic energy decreases if either the mass or the velocity of the object increases or if both increase.

Kinetic energy decreases if either the mass or the velocity of the object decreases or if both decrease.

Kinetic energy increases if either the mass or the velocity of the object decreases or if both decrease.

Answers

Kinetic energy decreases if either the mass or the velocity of the object decreases or if both decrease. and  Kinetic energy increases if either the mass or the velocity of the object increases or if both increase.

Explanation:

These statements explain the relationship between mass, velocity, and kinetic energy .i.e KE = ½mv², (3) Kinetic energy decreases if either the mass or the velocity of the object decreases or if b decrease. (4)Kinetic energy increases if either the mass or the velocity of the object decreases or if both decrease.

What is kinetic energy?

When a body has a mass of m kg and travels a velocity of v m/s then the kinetic energy is given by,

KE=1/2mv²

In the above relation, it is clearly shown that,

Kinetic energy is directly proportional to the mass of the body,

i.e

KE∝m

and also Kinetic energy directly varies with the square of the velocity of the body.

I.e

KE∝v²

Therefore, From the given expression we can conclude that when mass and velocity increased either one of them or both, the kinetic energy also gets increased.

And when mass and velocity decreased either one of them or both, the kinetic energy also gets decreased. Hence options (3) and (4) are correct.

To learn more about kinetic energy click:

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A child of mass 28 kg swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal, and the speed is 7 m/s. At this instant the cord is 3.30 m long. (Take the +x direction to be horizontal and to the right, the +y direction upward and +z direction to be out of the page.)

a. At this instant, what is the magnitude of the rate of change of the child's momentum?
b. At this instant, what is the (vector) net force acting on the child?

Answers

Answer:

Explanation:

(A)

Force is said to be the rate of change of momentum.

At the instant, the parallel component of the change of momentum is zero. SInce centripetal force always acts towards the center, hence the direction upwards ( along the positive y-axis).

[tex]\dfrac{dp}{dt} = F(+\hat y)[/tex]

[tex]= \dfrac{mv^2}{r}(+\hat y) \ \\ \\ = \dfrac{28 \ kg \times (7 \ m/s)^2}{3.3 \ m }( + \hat y) \\ \\ = \mathbf{416 \ N ( + \hat y)}[/tex]

(B)

The net force acting on the child is:

[tex]F_{net} = \dfrac{mv^2}{r}( + \hat y) \\ \\[/tex]

[tex]= \dfrac{28 \ kg \times (7 \ m/s)^2}{3.3 \ m }( + \hat y) \\ \\ = \mathbf{416 \ N ( + \hat y)}[/tex]

5 litres of alcohol have a mass of 4kg. calculate the density of alcohol in g/cm.​

Answers

Answer: 0.8 g/cm

Explanation:

p= m/V

= 4 kg/ 5 liter

= 0.8

Answer:

.8

Explanation:

give them brainliest

What happens in the crushing can experiment?

Answers

Explanation:

When hot water is poured on the can in a bucket of cold water, the can crushes off means it gets unshaped

A 0.5 kg mass on a spring undergoes simple harmonic motion with a total mechanical energy of 12 J. If the oscillation amplitude is 0.45 m, what is the frequency of the oscillation?

Answers

Answer:

The frequency of the oscillation is 2.45 Hz.

Explanation:

Given;

mass of the spring, m = 0.5 kg

total mechanical energy of the spring, E = 12 J

Determine the spring constant, k as follows;

E = ¹/₂kA²

kA² = 2E

k = (2E) / (A²)

k = (2 x 12) / (0.45²)

k = 118.519 N/m

Determine the angular frequency, ω;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{118.519}{0.5} } \\\\\omega = 15.396 \ rad/s[/tex]

Determine the frequency of the oscillation;

ω = 2πf

f = (ω) / (2π)

f = (15.396) / (2π)

f = 2.45 Hz

Therefore, the frequency of the oscillation is 2.45 Hz.

A long, uninsulated steam line with a diameter of 100 mm and a surface emissivity of 0.8 transports steam at 150°C and is exposed to atmospheric air and large surroundings at an equivalent temperature of 20°C. (a) Calculate the rate of heat loss per unit length for a calm day. (b) Calculate the rate of heat loss on a breezy day when the wind speed is 8

Answers

Answer:

Heat loss per unit length = 642.358 W/m

The heat loss per unit length on a breezy day during 8 m/s speed is = 1760.205 W/m

Explanation:

From the information given:

Diameter D [tex]= 100 mm = 0.1 m[/tex]

Surface emissivity ε = 0.8

Temperature of steam [tex]T_s[/tex] = 150° C = 423K

Atmospheric air temperature [tex]T_{\infty} = 20^0 \ C = 293 \ K[/tex]

Velocity of wind V = 8 m/s

To calculate average film temperature:

[tex]T_f = \dfrac{T_s+T_{\infty}}{2}[/tex]

[tex]T_f = \dfrac{423+293}{2}[/tex]

[tex]T_f = \dfrac{716}{2}[/tex]

[tex]T_f = 358 \ K[/tex]

To calculate volume expansion coefficient

[tex]\beta= \dfrac{1}{T_f} \\ \\ \beta= \dfrac{1}{358} \\ \\ \beta= 2.79 \times 10^{-3} \ K^{-1}[/tex]

From the table of "Thermophysical properties of gases at atmospheric pressure" relating to 358 K of average film temperature; the following data are obtained;

Kinematic viscosity (v) = 21.7984 × 10⁻⁶ m²/s

Thermal conductivity k = 30.608 × 10⁻³ W/m.K

Thermal diffusivity ∝ = 31.244 × 10⁻⁶ m²/s

Prandtl no. Pr = 0.698

Rayleigh No. for the steam line is determined as follows:

[tex]Ra_{D} = \dfrac{g \times \beta (T_s-T_{\infty}) \times D_b^3}{\alpha\times v}[/tex]

[tex]Ra_{D} = \dfrac{9.8 \times (2.79 *10^{-3})(150-20) \times (0.1)^3}{(31.244\times 10^{-6}) \times (21.7984\times 10^{-6})}[/tex]

[tex]Ra_{D} = 5.224 \times 10^6[/tex]

The average Nusselt number is:

[tex]Nu_D = \Big \{ 0.60 + \dfrac{0.387(Ra_D)^{1/6}}{[ 1+ (0.559/Pr)^{9/16}]^{8/27}} \Big \}^2[/tex]

[tex]Nu_D = \Big \{ 0.60 + \dfrac{0.387(5.224\times 10^6)^{1/6}}{[ 1+ (0.559/0.698)^{9/16}]^{8/27}} \Big \}^2[/tex]

[tex]Nu_D = \Big \{ 0.60 + \dfrac{5.0977}{[ 1.8826]^{8/27}}\Big \}^2[/tex]

[tex]Nu_D = \Big \{ 0.60 + 4.226 \Big \}^2[/tex]

[tex]Nu_D = 23.29[/tex]

However, for the heat transfer coefficient; we have:

[tex]h_D = \dfrac{Nu_D\times k}{D_b} \\ \\ h_D = \dfrac{(23.29) \times (30.608 \times 10^{-3} )}{0.1}[/tex]

[tex]h_D = 7.129 \ Wm^2 .K[/tex]

Hence, Stefan-Boltzmann constant [tex]\sigma = 5.67 \times 10^{-8} \ W/m^2.K^4[/tex]

Now;

To determine the heat loss using the formula:

[tex]q'_b = q'_{ev} + q'_{rad} \\ \\ q'_b = h_D (\pi D_o) (T_t-T_{\infty})+\varepsilon(\pi D_b)\sigma (T_t^4-T_{\infty }^4)[/tex]

[tex]q'_b = (7.129)(\pi*0.1) (423-293) + (0.8) (\pi*0.1) (5.67 *10^{-8}) (423^4-293^4) \\ \\ q'_b = 291.153 + 351.205 \\ \\ \mathbf{q'_b = 642.258 \ W/m}[/tex]

Now; here we need to determine the Reynold no and the average Nusselt number:

[tex]Re_D = \dfrac{VD_b}{v } \\ \\ Re_D = \dfrac{8 *0.1}{21.7984 \times 10^{-6}} \\ \\ Re_D = 3.6699 \times 10^4[/tex]

However, to determine the avg. Nusselt no by using Churchill-Bernstein correlation, we have;

[tex]Nu_D = 0.3 + \dfrac{0.62 \times Re_D^{1/2}* Pr^{1/3}}{[1+(0.4/Pr)^{2/3}]^{1/4}} [1+ (\dfrac{Re_D}{282000})^{5/8}]^{4/5}[/tex]

[tex]Nu_D = 0.3 + \dfrac{0.62 \times (3.6699*10^4)^{1/2}* (0.698)^{1/3}}{[1+(0.4/0.698)^{2/3}]^{1/4}} [1+ (\dfrac{3.669*10^4}{282000})^{5/8}]^{4/5}[/tex]

[tex]Nu_D = (0.3 +\dfrac{105.359}{1.140}\times 1.218) \\ \\ Nu_D = 112.86[/tex]

SO, the heat transfer coefficient for forced convection is determined as follows afterward:

[tex]h_D = \dfrac{Nu_{D}* k}{D_b} \\ \ h_D = \dfrac{112.86*30.608 *10^{-3}}{0.1} \\ \\ h_D = 34.5 \ W/m^2 .K[/tex]

Finally; The heat loss per unit length on a breezy day during 8 m/s speed is:

[tex]q'b = h_D (\pi D_b) (T_s-T_{\infty}) + \varepsilon (\pi D_b) \sigma (T_s^4-T_ {\infty}^4) \\ \\ q'b = (34.5) (\pi *0.1) (423-293) + (0.8) (\pi*0.1) (5.67*10^{-8}) (423^4 - 293^4) \\ \\ = 1409 +351.205 \\ \\ \mathbf{q'b = 1760.205 \ W/m}[/tex]

What is it called when the moon passes through the penumbra of Earth’s shadow?(1 point)

total lunar eclipse

total solar eclipse

partial lunar eclipse

partial solar eclipse

Answers

Answer:  I'm not sure, but I think it would be a total lunar eclipse

When the moon passes through the penumbra of Earth’s shadow it is referred to as partial lunar eclipse. The correct option is C.

What is partial lunar eclipse?

A partial lunar eclipse occurs when the moon is not completely immersed in the umbra of the earth's shadow.

During a partial solar eclipse, the Moon, Sun, and Earth do not align perfectly straight, and the Moon casts only the penumbra of its shadow on Earth. From our vantage point, it appears that the Moon has eaten the Sun.

A shadow's penumbra is the lighter outer edge. Partial solar eclipses are caused by the Moon's penumbra, while penumbral lunar eclipses are caused by the Earth's penumbra. The penumbra is a type of lighter shadow.

The Penumbra is a half-shadow region that occurs when an object only partially covers a light source.

Thus, the correct option is C.

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what is the distance a train can travel if its speed is 20mph over a time of 5.6 hours (show all 3 steps)

Answers

Answer:

distance = 112 miles

Explanation:

its 12 miles every 0.6 in a hour

Which is an example of a noncontract force?
(A) elastic force
(B) Normal force
(C)Applied force
(D) Electric Force

Answers

Answer:

electric force

Explanation:

its a contact and noncontact force

Mars orbits the Sun in 1.87 Earth years. How far is Mars from the Sun?

Answers

Answer:

151.12 million miles i actually just did this question on a test

Explanation:

Mars orbits—or completes one revolution—around the Sun every 686.98 Earth days, or once every 1.88 Earth years.

What is meant by one revolution?

The term "revolution" describes how an item moves in its orbit around another object. For instance, the 24-hour day is created as the Earth rotates on its own axis.

The 365-day year is made possible by the Sun's rotation of the Earth. A planet spins around a satellite.

While orbiting the Sun, Mars travels at an average speed of 53,979 miles per hour, which equates to 86,871 km per hour.

Thus, Mars orbits—or completes one revolution—around the Sun every 686.98 Earth days, or once every 1.88 Earth years.

To learn more about One revolution, refer to the below link:

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Cells use nutrients and oxygen to supply the body with the energy it needs. What three-body systems are working together in this situation?
A
nervous, digestive, and circulatory systems

B
digestive, circulatory, and excretory systems

C
circulatory, immune, and respiratory systems

D
digestive, respiratory, and circulatory systems

Answers

Answer:

The respiratory system provides oxygen for cells, while the circulatory system transports oxygen to cells.

Explanation:

so the answer is D

(will give brainliest to whoever is correct and shows reasoning) What is the acceleration of an object that has a velocity of 60m/s and is moving in a circle of radius 50m?

Answers

Answer:

5.0/s

Explanation:

Answer:

b and a it is this that abewsr

A large wooden block of weight 30n, is observed to push down a ramp. The block is pushed down an inclined plane as shown below and the applied force is parallel to the plane and has a magnitude of 5 n as shown. The force of kinetic friction is 17n the block is pushed 6m down along the inclination of the ramp is 30 degrees with the horizontal

Answers

Answer:

OMG IM ON THE SAME QUESTION

Explanation:

Lil' Ricky is trying on his Halloween costume in front of a flat (plane) mirror. Lil Ricky stands 1.5 m from the mirror.
Which statement correctly describes the image formed in the mirror?
A)
It is upright and 1.5 m behind the mirror.
B)
It is upright and 1.0 m behind the mirror.
C)
It is inverted and 1.5 m behind the mirror.
D)
It is inverted and 1.5 m in front of the mirror.

Answers

Answer:

The correct answer is A) It is upright and 1.5m behind the mirror

Explanation:

Your reflection must be upright, meaning vertical/erect, and the distance will be the exact same. Also, the reflected ray appears as if it had traveled from an object located behind the mirror.

A physics professor wants to demonstrate the large size of the henry unit. On the outside of a 16-cm-diameter plastic hollow tube, she wants to wind an air-filled solenoid with self-inductance of 1.0 H using copper wire with a 0.79-mm diameter. The solenoid is to be tightly wound with each turn touching its neighbor (the wire has a thin insulating layer on its surface so the neighboring turns are not in electrical contact).

Required:
a. How long will the plastic tube need to be?
b. How many kilometers of copper wire will be required?
c. What will be the resistance of this solenoid?

Answers

Answer:

a) the plastic tube need to be 24.7 m long

b) the kilometer of copper wire required is 15.7

c) the resistance of this solenoid is 5538.8 2 ohms

Explanation:

Given the data in the question;

we determine the length of the plastic tube. assuming the solenoid is long.

the self inductance of a long solenoid is;

L = μ₀n²πr²l

μ₀ = 4π × 10⁻⁷ T-m/A

where

n = number of turns per unit length

r = radius of the solenoid = 8cm (as the diameter of the plastic hollow tube is 16 cm)

l = length of the solenoid or the length of the plastic tube

we find n = number of turns per unit length

given that, the copper wire to be wound around the solenoid is 0.79 mm in diameter

number of turns per meter = n = 1 / ( 0.79 × 10⁻³ m ) = 1265.8 turns/meter

So from our previous formula, we find l

L = μ₀n²πr²l

we substitute

1.0 H = (4π × 10⁻⁷ T-m/A)( 1265.8 )²(3.14)(0.08)² ( l)

1 = 0.04048 × l

l = 1 / 0.04048

l = 24.7 m

Therefore, the plastic tube need to be 24.7 m long

b)  

n = number of turns per unit length = 1265.8 turns/metre

so, the length of the plastic tube over which the copper wire is to be wound,

number of turns of copper wire required = n × l

= 1265.8 turns/meter × 24.7 m

= 31,265.26 turns

Now each turn of the copper wire is to be wound across the 18cm diameter of the plastic tube.

so for each turn length of copper wire required = 2π × r

= 2π × 0.08 m

= 0.5026548 m

So copper wire required for 31,265.26 turns will be;

⇒ 31,265.26 × 0.5026548 = 15715.63m = 15.7 km

Therefore, the kilometer of copper wire required is 15.7

c)

p = resistivity of copper = 1.68 × 10⁻⁸ ohm-m

Resistance = pl/a

where l is length of copper wire, a is cross sectional area;

diameter of copper wire is 0.79-mm

radius of copper wire is 0.79/2 = 0.395 mm = 0.000395 m

area of cross section of copper wire a = πr² = π( 0.00395)² = 4.9 × 10⁻⁷ m²

Resistance = pl/a

we substitute

Resistance = [(1.68 × 10⁻⁸ ohm-m)( 15715.63m )] / [ 4.9 × 10⁻⁷ m² ]

Resistance =  5538.8 2 ohms

Therefore,  the resistance of this solenoid is 5538.8 2 ohms

The water line from the street to my house is 1 inch diameter and made of PVC (i.e. smooth). The line is roughly 450 ft long. The water pressure in the line at the street is 130 psig. If I flow 10 gpm through the pipe, what pressure would I expect when I get to my house. My house is 10 ft higher in elevation than the water line at the street

Answers

Answer:

The right solution is "126 Psi".

Explanation:

The given values are:

P₁ = 130 psig

i.e.,

   = [tex]130\times 6.894[/tex]

   = [tex]896.22 \ Kpa[/tex]

or,

   = [tex]896.22\times 10^3 \ Pa[/tex]

Z₂ = 10ft

    = 3.05 m

[tex]\delta[/tex] = 1000 kg/m³

According to the question,

Z₁ = 0

V₁ = V₂

As we know,

⇒  [tex]\frac{P_1}{\delta_g} +\frac{V_1^2}{2g} +Z_1=\frac{P_2}{\delta_g} +\frac{V_2^2}{2g} +Z_2[/tex]

On substituting the values, we get

⇒  [tex]\frac{P_1}{\delta_g} +0+0=\frac{P_2}{\delta_g} +0+Z_2[/tex]

⇒  [tex]\frac{896.22\times 10^3}{1000\times 9.8} =\frac{P_2}{1000\times 9.8} +3.05[/tex]

⇒  [tex]P_2=866330 \ P_a[/tex]

i.e.,

⇒       [tex]=866330\times 0.000145[/tex]

⇒       [tex]=126 \ Psi[/tex]

A block is sliding along a horizontal surface with 100 J of kinetic energy. The kinetic frictional force stops the block in 5 seconds.

What is the rate (in J/s) at which the force of kinetic friction does work?

Answers

Answer:

20 J/s

Explanation:

Given data

Kinetic energy=100 J

Time= 5 seconds

Hence the rate at which the kinetic friction work is

=100/5

=20 J/s

Therefore the answer is 20 J/s

a sharp image is formed when light reflects from a

Answers

Answer:

Regular reflection

Explanation:

Regular reflection occurs when light reflects off a very smooth surface and forms a clear image.

i hope this helps a bit.

According to the context, a sharp image is formed when light reflects from a regular reflection.

What is regular reflection?

It is reflection without diffusion that obeys the laws of geometrical optics, as in mirrors.

This reflection of light happens when the angles that the two rays determine with the surface are equal.

Therefore, we can conclude that according to the context, a sharp image is formed when light reflects from a regular reflection.

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Would sound travel faster in an oven or a freezer?

Answers

Answer:

An Oven

Explanation:

The heat is higher, so it moves faster. Shile in a freezer the particles are extremely slow!

A box with a mass of 0.5 slugs lies on an inclined plane making an angle of 30o with the horizontal. If the coefficient of kinetic friction between the box and the plane is 0.4, what is the magnitude of the force that must be applied parallel to the incline to keep the box moving down the incline at a constant speed

Answers

Answer: 10.98 N

Explanation:

Given

mass of box is [tex]m=0.5\ slugs\approx 7.3\ kg[/tex]

The coefficient of kinetic friction is [tex]\mu= 0.4[/tex]

It is given that on the application of force box started moving with constant speed i.e. there is no net external force

[tex]\Rightarrow F+mg\sin 30^{\circ}=\mu mg\cos 30^{\circ}\\\Rightarrow F=mg(\mu \cos 30^{\circ}-\sin 30^{\circ})\\\Rightarrow F=7.3\times 9.8(0.866\times0 .4-0.5)=-10.98\ N\\[/tex]

The negative sign indicates direction of force is opposite i.e. it must be applied upwards

Linearly polarized light whose Jones vector is [0 1] (horizontally polarized) is sent through a train of two linear polarizers. The first is oriented with its transmission axis at 45 degrees and the second has its transmission axis vertical. Show that the emerging light is linearly polarized in the vertical direction; that is, the plane of polarization has been rotated by 90 degrees.

Answers

Answer:

Following are the solution to the given question:

Explanation:

The input linear polarisation was shown at an angle of [tex]2 \mu[/tex]. It's a very popular use of a half-wave plate. In particular, consider the case [tex]\mu = 45 \pm[/tex], at which the angle of rotation is [tex]90\pm[/tex]. HWP thereby provides a great way to turn, for instance, a linear polarised light that swings horizontally to polarise vertically. Illustration of action on event circularly polarized light of the half-wave platform. Customarily it is the slow axis of HWP that corresponds to either the rotation. Note that perhaps the vector of polarization is "double-headed," i.e., the electromagnetic current swinging back and forward in time. Therefore the turning angle could be referred to as the rapid axis to reach the same result. Please find the attached file.

Find all of those elements on the periodic table

Answers

Answer:

hydrogen, helium, lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine, neon, sodium, magnesium, aluminum, silicon, phosphorus, sulfur, chlorine, argon, potassium, calcium, scandium, titanium, vanadium, chromium, maganese, iron, cobalt, nickel, copper, zinnc, gallium, germanium, aresnic, selenium, bromine, krypton, rubidium, strongtium, yttrium, zirconium, niobium, molybdenum, technetium, ruthenium, rhodium, palladium, silver, cadmium, indium, tin, atimony, tellurium, iodine, xenon, cesium, barium, lanthanum, cerium, praseodyumium, neodymim, promethium, samarium, europium, gadolinium, terbium, dysprosium, holmium, erbium, thulium, ytterbium, luteium, hafnium, tantalum, tungsten, rhenium, osmium, irdium, platinum, gold, mercury, thallium, lead bismuth, polnium, astatine, radon, rancium, radium, actinum, thorium, protactinium

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