QUESTION 27
The titanium shell of an SR-71 airplane would expand when flying at a speed exceeding 3 times the speed of sound. If the skin of the
plane is 400 degrees C and the linear coefficient of expansion for titanium is 5x10-6/C when flying at 3 times the speed of sound, how
much would a 10-meter long (originally at oC) portion of the airplane expand? Write your final answer in centimeters and show all of your
work.

Answers

Answer 1

Answer:

2 cm.

Explanation:

Data obtained from the question include the following:

Original Length (L₁ ) = 10 m

Initial temperature (T₁) = 0°C

Final temperature (T₂) = 400°C

Linear expansivity (α) = 5×10¯⁶ /°C

Increase in length (ΔL) =..?

Next, we shall determine the temperature rise (ΔT).

This can be obtained as follow:

Initial temperature (T₁) = 0°C

Final temperature (T₂) = 400°C

Temperature rise (ΔT) =..?

Temperature rise (ΔT) = T₂ – T₁

Temperature rise (ΔT) = 400 – 0

Temperature rise (ΔT) = 400°C

Thus, we can obtain the increase in length of the airplane by using the following formula as illustrated below:

Linear expansivity (α) = increase in length (ΔL) /Original Length (L₁ ) × Temperature rise (ΔT)

α = ΔL/(L₁ × ΔT)

Original Length (L₁ ) = 10 m

Linear expansivity (α) = 5×10¯⁶ /°C

Temperature rise (ΔT) = 400°C

Increase in length (ΔL) =..?

α = ΔL/(L₁ × ΔT)

5×10¯⁶ = ΔL/(10 × 400)

5×10¯⁶ = ΔL/4000

Cross multiply

ΔL = 5×10¯⁶ × 4000

ΔL = 0.02 m

Converting 0.02 m to cm, we have:

1 m = 100 cm

Therefore, 0.02 m = 0.02 × 100 = 2 cm.

Therefore, the length of the plane will increase by 2 cm.


Related Questions

A current of 5 A is flowing in a 20 mH inductor. The energy stored in the magnetic field of this inductor is:_______

a. 1J.
b. 0.50J.
c. 0.25J.
d. 0.
e. dependent upon the resistance of the inductor.

Answers

Answer:

C. 0.25J

Explanation:

Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;

L is the inductance

I is the current flowing in the inductor

Given parameters

L = 20mH = 20×10^-3H

I = 5A

Required

Energy stored in the magnetic field.

E = 1/2 × 20×10^-3 × 5²

E = 1/2 × 20×10^-3 × 25

E = 10×10^-3 × 25

E = 0.01 × 25

E = 0.25Joules.

Hence the energy stored in the magnetic field of this inductor is 0.25Joules

You have a horizontal grindstone (a disk) that is 95 kg, has a 0.38 m radius, is turning at 87 rpm (in the positive direction), and you press a steel axe against the edge with a force of 16 N in the radial direction.
(a) Assuming the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone.
(b) How many turns will the stone make before coming to rest?

Answers

Answer:

Explanation:

The moment of inertia of the disk I  = 1/2 m R² where R is radius of the disc and m is its mass .

putting the values

I = .5 x 95 x .38²

= 6.86 kg m²

n = 87 rpm = 87 / 60 rps

n = 1.45 rps

angular velocity ω = 2π n , n is frequency of rotation .

= 2 x 3.14 x 1.45

= 9.106 radian /s

frictional force = 16 x .2

= 3.2 N

torque created by frictional force = 3.2 x .38

= 1.216 N.m

angular acceleration = torque / moment of inertia

= - 3.2 / 6.86

α  = - 0.4665 rad /s²

b ) ω² = ω₀² +  2 α θ , where α is angular acceleration

0 = 9.106² - 2 x .4665 θ

θ = 88.87 radian

no of turns = 88.87 / 2π

= 14.15  turns

"A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if"

Answers

Answer:

A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if

the dispersion is great

The physics of wind instruments is based on the concept of standing waves. When the player blows into the mouthpiece, the column of air inside the instrument vibrates, and standing waves are produced. Although the acoustics of wind instruments is complicated, a simple description in terms of open and closed tubes can help in understanding the physical phenomena related to these instruments. For example, a flute can be described as an open-open pipe because a flutist covers the mouthpiece of the flute only partially. Meanwhile, a clarinet can be described as an open-closed pipe because the mouthpiece of the clarinet is almost completely closed by the reed.
1. Consider a pipe of length 80.0 cm open at both ends. What is the lowest frequency f of the sound wave produced when you blow into the pipe?
2. A hole is now drilled through the side of the pipe and air is blown again into the pipe through the same opening. The fundamental frequency of the sound wave generated in the pipe is now:______.
a. the same as before.
b. lower than before.
c. higher than before.
3. If you take the original pipe in Part A and drill a hole at a position half the length of the pipe, what is the fundamental frequency of the sound that can be produced in the pipe?
4. What frequencies, in terms of the fundamental frequency of the original pipe in Part A, can you create when blowing air into the pipe that has a hole halfway down its length?
4-1. Recall from the discussion in Part B that the standing wave produced in the pipe must have an antinode near the hole. Thus only the harmonics that have an antinode halfway down the pipe will still be present.
A. Only the odd multiples of the fundamental frequency.
B. Only the even multiples of the fundamental frequency.
C. All integer multiples of the fundamental frequency.
E. What length of open-closed pipe would you need to achieve the same fundamental frequency as the open pipe discussed in Part A?
A. Half the length of the open-open pipe.
B. Twice the length of the open-open pipe.
C. One-fourth the length of the open-open pipe.
D. Four times the length of the open-open pipe.
E. The same as the length of the open-open pipe.
F. What is the frequency of the first possible harmonic after the fundamental frequency in the open-closed pipe described in Part E?
F-1. Recall that possible frequencies of standing waves that can be generated in an open-closed pipe include only odd harmonics. Then the first possible harmonic after the fundamental frequency is the third
harmonic.

Answers

Answer:

1) f = 214 Hz , 2)  answer is c , 3) f = 428 Hz , 4)   f₂ = 428 Hz ,   f₃ = 643Hz

Explanation:

1) A tube with both ends open, the standing wave has a maximum amplitude and a node in its center, therefore

                L = λ / 2

               λ  = 2L

               λ  = 2 0.8

               λ  = 1.6 m

wavelength and frequency are related to the speed of sound (v = 343 m / s)

                v =λ  f

                f = v / λ  

                f = 343 / 1.6

                f = 214 Hz

2) In this case the air comes out through the open hole, so we can assume that the length of the tube is reduced

           λ' = 2 L ’

          as L ’<L₀

          λ' <λ₀

          f = v / λ'

          f' > fo

the correct answer is c

3) in this case the length is L = 0.40 m

          λ = 2 0.4 = 0.8 m

          f = 343 / 0.8

          f = 428 Hz

4) the different harmonics are described by the expression

         λ = 2L / n           n = 1, 2, 3

         λ₂ = L

         f₂ = 343 / 0.8

         f₂ = 428 Hz

         λ₃ = 2 0.8 / 3

         λ₃ = 0.533 m

         f₃ = 343 / 0.533

         f₃ = 643 Hz

4,1) as we have two maximums at the ends, all integer multiples are present

       the answer is C

E) the length of an open pipe created that has a wavelength of lam = 1.6 m is requested

in this pipe there is a maximum in the open part and a node in the closed part, so the expression

        L = λ / 4

        L = 1.6 / 4

        L = 0.4 m

the answer is C

F) in this type of pipe the general expression is

           λ = 4L / n         n = 1, 3, 5 (2n + 1)

therefore only odd values ​​can produce standing waves

           λ₃ = 4L / 3

           λ₃ = 4 0.4 / 3

           λ₃ = 0.533

           f₃ = 343 / 0.533

           f₃ = 643 Hz

Mars Rover When the Mars rover was deployed on the surface of Mars in July 1997, radio signals took about 12 minmin to travel from Earth to the rover.
How far was Mars from Earth at that time?

Answers

Answer:

s = 2.16 x 10¹¹ m

Explanation:

Since, the waves travelling from Earth to the Mars rover are electromagnetic. Therefore, there speed must be equal to the speed of light. So, from the equation given below:

s = vt

where,

s = the distance between Earth and Mars = ?

v = speed of the wave = speed of light = 3 x 10⁸ m/s

t = time taken by the radio signals to reach the rover from Earth

t = (12 min)(60 s/1 min) = 720 s

Therefore,

s = (3 x 10⁸ m/s)(720 s)

s = 2.16 x 10¹¹ m

What is the speed of light (in m/s) in air? (Enter your answer to at least four significant figures. Assume the speed of light in a vacuum is 2.997 ✕ 108 m/s.) m/s What is the speed of light (in m/s) in polystyrene? m/s

Answers

Answer:

The speed of light in air is 2.996x10⁸ m/s, and polystyrene is 1.873x10⁸ m/s.

Explanation:

To find the speed of light in air and in polystyrene we need to use the following equation:

[tex] c_{m} = \frac{c}{n} [/tex]

Where:

[tex]c_{m}[/tex]: is the speed of light in the medium

n: is the refractive index of the medium

In air:

[tex]c_{a} = \frac{c}{n_{a}} = \frac{2.997 \cdot 10^{8} m/s}{1.0003} = 2.996 \cdot 10^{8} m/s[/tex]

In polystyrene:

[tex]c_{p} = \frac{c}{n_{p}} = \frac{2.997 \cdot 10^{8} m/s}{1.6} = 1.873 \cdot 10^{8} m/s[/tex]  

Therefore, the speed of light in air is 2.996x10⁸ m/s, and polystyrene is 1.873x10⁸ m/s.

I hope it helps you!

An interference pattern is produced by light with a wavelength 590 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.580 mm .


Required:

a. If the slits are very narrow, what would be the angular position of the first-order, two-slit, interference maxima?

b. What would be the angular position of the second-order, two-slit, interference maxima in this case?

Answers

Answer:

a. 0.058°

b.  0.117°

Explanation:

a. The angular position of the first-order is:

[tex] d*sin(\theta) = m\lambda [/tex]

[tex] \theta = arcsin(\frac{m \lambda}{d}) = arcsin(\frac{1* 590 \cdot 10^{-9} m}{0.580 \cdot 10^{-3} m}) = 0.058 ^{\circ} [/tex]

Hence, the angular position of the first-order, two-slit, interference maxima is 0.058°.

b. The angular position of the second-order is:

[tex] \theta = arcsin(\frac{m \lambda}{d}) = arcsin(\frac{2* 590 \cdot 10^{-9} m}{0.580 \cdot 10^{-3} m}) = 0.12 ^{\circ} [/tex]

Therefore, the angular position of the second-order, two-slit, interference maxima is 0.117°.

I hope it helps you!

The left end of a long glass rod 8.00 cm in diameter and with an index of refraction of 1.60 is ground and polished to a convex hemispherical surface with a radius of 4.00 cm. An object in the form of an arrow 1.70 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the convex surface.
A) Find the position of the image of the arrow formed by paraxial rays incident on the convex surface.
B) Find the height of the image formed by paraxial rays incident on the convex surface.
C) Is the image erect or inverted?

Answers

Answer:

A) 0.1477

B) 0.65388 mm

C) object is inverted

Explanation:

The formula for object - image relationships for spherical reflecting surface is given as;

n1/s + n2/s' = = (n2 - n1)/R

Where;

n1 & n2 are the Refractive index of both surfaces

s is the object distance from the vertex of the spherical surface

s' is the image distance from the vertex of the spherical surface

R is the radius of the spherical surface

We are given;

index of refraction of glass; n2 = 1.60

s = 24 cm = 0.24 m

R = 4 cm = 0.04 m

index of refraction of air has a standard value of 1. Thus; n1 = 1

a) So, making s' the subject from the initial equation, we have;

s' = n2/[((n2 - n1)/R) - n1/s]

Plugging in the relevant values, we have;

s' = 1.6/[((1.6 - 1)/0.04) - 1/0.24]

s' = 0.1477

b) The formula for lateral magnification of spherical reflecting surfaces is;

m = -(n1 × s')/(n2 × s) = y'/y

Where;

m is the magnification

n1, n2, s & s' remain as earlier explained

y is the height of the object

y' is the height of the image

Making y' the subject, we have;

y' = -(n1 × s' × y)/(n2 × s)

We are given y = 1.7 mm = 0.0017 m and all the other terms remain as before.

Thus;

y' = -(1 × 0.1477 × 0.0017)/(1.6 × 0.24)

y' = - 0.00065388021 m = -0.65388 mm

C) since y' is negative and y is positive therefore, m = y'/y would result in a negative value.

Now, in object - image relationships for spherical reflecting surface, when magnification is positive, it means the object is erect and when magnification is negative, it means the object is inverted.

Thus, the object is inverted since m is negative.

A stonecutter's chisel has an edge area of 0.7 cm2. If the chisel is struck with a force of 42 N, what is the pressure exerted on the stone

Answers

Answer:

The pressure is [tex]P = 583333 \ N/m^2[/tex]

Explanation:

From the question we are told that

  The area of the edge is  [tex]A = 0.72 cm^2 = 0.72 *10^{-4}\ m[/tex]

    The  force is [tex]F = 42 \ N[/tex]

The pressure is mathematically represented as

            [tex]P = \frac{F}{A}[/tex]

substituting values

           [tex]P = \frac{42}{0.72*10^{-4}}[/tex]

           [tex]P = 583333 \ N/m^2[/tex]

beam of white light goes from air into water at an incident angle of 58.0°. At what angles are the red (660 nm) and blue (470 nm) parts of the light refracted? (Enter your answer to at least one decimal place.) red ° blue °

Answers

Answer:

For red light= 39.7°

Blue light 39.2°

Explanation:

Given that refractive index for red light is 1.33 and that of blue light is 1.342

So angle of refraction for red light will be

Sinစi/ (sinစ2) =( nw)r/ ni

Sin 58° x 1.000293/1.33. =( sinစ2)r

0.64= sinစ2r

Theta2r = 39.7°

For blue light

Sinစi/ (sinစ2) =( nw)b/ ni

Sin 58° x 1.000293/1.342 =( sinစ2)b

0.632= sinစ2r

Theta2b= 39.19°

a solenoid that is 98.6 cm long has a cross-sectional area of 24.3 cm2. There are 1310 turns of a wire carrying a current of

Answers

Complete question:

A solenoid that is 98.6 cm long has a cross-sectional area of 24.3 cm2. There are 1310 turns of a wire carrying a current of 6.75 A. (a) Calculate the energy density of the magnetic field inside the solenoid. (b) Find the total energy stored in the magnetic field there (neglect end effects).

Answer:

(a) the energy density of the magnetic field inside the solenoid is 50.53 J/m³

(b) the total energy stored in the magnetic field is 0.121 J

Explanation:

Given;

length of the solenoid, L = 98.6 cm = 0.986 m

cross-sectional area of the solenoid, A = 24.3 cm² = 24.3 x 10⁻⁴ m²

number of turns of the solenoid, N = 1310 turns

The magnitude of the magnetic field inside the solenoid is given by;

B = μ₀nI

B = μ₀(N/L)I

Where;

μ₀ is permeability of free space, = 4π x 10⁻⁷ m/A

[tex]B = \frac{4\pi*10^{-7}*1310*6.75}{0.986} \\\\B = 0.01127 \ T[/tex]

(a) Calculate the energy density of the magnetic field inside the solenoid

[tex]u = \frac{B^2}{2 \mu_o}\\\\u = \frac{(0.01127)^2}{2*4\pi *10^{-7}} \\\\u = 50.53 \ J/m^3[/tex]

(b) Find the total energy stored in the magnetic field

U = uV

U = u (AL)

U = 50.53 (24.3 x 10⁻⁴  x 0.986)

U = 0.121 J

A scuba diver fills her lungs to capacity (6.0 L) when 10.0 m below the surface of the water and begins to ascend to the surface. Assume the density of the water in which she is swimming is 1000 kg/m3 and use g = 10 m/s2A. Assuming the temperature of the air in her lungs is constant, to what volume must her lungs expand when she reaches the surface of the water?B. What effect would the warming of the air in her lungs have on the volume needed when she surfaces?C. Assuming the temperature of the air in her lungs is constant, what effect does her ascent have on the vrms of the air molecules in her lungs?

Answers

Answer:

Explanation:

As temperature is constant , we shall apply Boyle's law

P₁V₁ = P₂V₂

P₁ = pressure at depth of 10 m

= P + hdg , h = 10 , d = 10³ , g = 10

P is atmospheric pressure which is 10⁵ Pa

P₁ = 10⁵ + 10 x 10³ x 10

= 2 x 10⁵

applying the formula

2 x 10⁵ x 6 = 10⁵ x v

v = 2 x 6 = 12 L

volume will be doubled at the surface .

B )

warming of air at the surface will increase the volume of air in her lungs so so she will need more lung capacity .

C )

The rms value of a gas depends upon the temperature of the gas . As temperature of the gas is constant , the rms value of the gas particles will remain constant when she goes to the surface .

The lungs will expand 12 L when she reaches the surface of the water, and the warming of the air results in more lung capacity, and [tex]\rm V_{rms}[/tex] the value remains the same.

What is Boyle's law?

According to the law, the pressure of the gas is inversely proportional to the volume of the gas. In other words when the pressure of the gas increases the volume of the gas decreases.

We know the pressure at the 10 meters depth:

[tex]\rm P_1 = P+h\times \rho\times g[/tex]

Where P = Atmospheric pressure

            h = Depth

            ρ =Density of the water

We have: [tex]\rm P = 10^5 \ Pa[/tex], h = 10 meters, and [tex]\rm \rho = 1000 \ kg/m^3[/tex], and [tex]\rm g = 10 \ m/s^2[/tex]

Putting the values in the above equation, we get:

[tex]\rm P_1 = 10^5+ 10\times 1000\times 10[/tex]

[tex]\rm P_1 = 2\times 10^5[/tex]

From the Boyle's law:

[tex]\rm P_1\times V_1 = P_2\times V_2[/tex]

[tex]\rm 2\times10^5\times 6 = 10^5\times V_2[/tex]

[tex]\rm V_2 = 12 \ L[/tex]

We know that as the air at the surface warms, the volume of air in her lungs expands, requiring more lung capacity.

The temperature of the gas is constant and [tex]\rm V_{rms}[/tex] values for gas depend on the temperature of the gas, but here the temperature of the gas is constant thus, the  [tex]\rm V_{rms}[/tex] will remains constant.

Thus, the lungs will expand 12 L when she reaches the surface of the water, and the warming of the air results in more lung capacity, and [tex]\rm V_{rms}[/tex] the value remains the same.

Learn more about the Boyle's law here:

https://brainly.com/question/23715689

A person of 70 kg standing on an un-deformable horizontal surface. She bends her knees and jumps up from rest, achieving a launching speed of 1.7 m/s. The launching process lasts 0.1 second. Calculate the average force exerted by the surface on the person during the launch.

Answers

Answer:

1190 N

Explanation:

Force: This can be defined as the product of mass and velocity. The unit of force is Newton(N).

From the question,

F = ma................. Equation 1

Where F = average force, m = mass, a = acceleration.

But,

a = (v-u)/t................ Equation 2

Where v = final velocity, u = initial velocity, t = time.

Substitute equation 2 into equation 1

F = m(v-u)/t.............. Equation 3

Given: m = 70 kg, v = 1.7 m/s, u = 0 m/s (from rest), t = 0.1 s.

Substitute into equation 3

F = 70(1.7-0)/0.1

F = 1190 N.

An object on a level surface experiences a horizontal force of 12.7 N due to kinetic friction. The coefficient of kinetic friction is 0.42.
What is the mass of the object? (Express your answer to two significant figures)kg

Answers

Answer:

The mass of the object is 3.08 kg.

Explanation:

The horizontal force is12.7 N and the coefficient of the kinetic fraction are 0.42. Now we have to compute the mass of the object. Thus, use the below formula to find the mass of the object.

Let the mass of the object = m.

The coefficient of kinetic friction, n = 0.42

Therefore,  

Force, F = n × mg

12.7 = 0.42 × 9.8 × m

m = 3.08 kg

The mass of the object is 3.08 kg.

A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N, what is the fundamental frequency

Answers

Answer:

f₀ = 158.12 Hertz

Explanation:

The fundamental frequency of the string  f₀ is expressed as f₀ = V/4L where V is the speed experienced by the string.

[tex]V = \sqrt{\frac{T}{\mu} }[/tex] where T is the tension in the string and  [tex]\mu[/tex] is the density of the string

Given T = 600N and [tex]\mu[/tex] = 0.015 g/cm  = 0.0015kg/m

[tex]V = \sqrt{\frac{600}{0.0015} }\\ \\V = \sqrt{400,000}\\ \\V = 632.46m/s[/tex]

The next is to get the length L of the string. Since the string is stretched and fixed at both ends, 200 cm apart, then the length of the string in metres is 2m.

L = 2m

Substituting the derived values into the formula f₀ = V/2L

f₀ = 632.46/2(2)

f₀ = 632.46/4

f₀ = 158.12 Hertz

Hence the fundamental frequency of the string is 158.12 Hertz

Question 5 of 10
Heat is being transferred through currents within a liquid. When will this heat
transfer mostly end?
O A. When the substance changes state and becomes a gas
O B. When the entire liquid is a single temperature
O C. When the substance is very hot on top and cold beneath
O D. When the particles stop bumping into each other
SUBMIT

Answers

Answer:

When the entire liquid is a single temperature

Explanation:

When a liquid is heated, a convection current is set up. Convection is the movement of

fluid particles in response to a temperature gradient.

When you start heating a liquid, the particles near the base of the heating vessel increase in temperature, become less dense and rise upwards while the denser particles move downwards. This convection current will continue until an equilibrium temperature is obtained throughout the liquid.

What is the smallest value of n for which the wavelength of a Balmer series line is less than 400 nm

Answers

Answer:

The smallest value is n= 2

Explanation:

The balmer equation is given below

1/λ = R(1/4 - 1/n₂²).

R= 1.0973731568508 × 10^7 m^-1

λ= 400*10^-9 m

(400*10^-9)= 1.0973731568508 × 10^7 (1/4-1/n²)

(400*10^-9)/1.0973731568508 × 10^7

= 1/4 - 1/n²

364.51 *10^-16= 1/4 - 1/n²

1/n²= 1/4 -364.51 *10^-16

1/n² = 0.25-3.6451*10^-14

1/0.25= n²

4= n²

√4= n

2= n

The smallest value is N= 2

Which of these cannot be a resistor in a series or parallel circuit?
A)switch
B) battery
C) light bulb
D) all of these are resistors

Answers

Answer:

it is going to D. all of these are resistors

Inside the wall of a house, an L-shaped section of hot-water pipe consists of three parts: a straight, horizontal piece 28.0 cm long, an elbow, and a straight vertical piece ℓ = 159 cm long. A stud and a second-story floorboard hold the ends of this section of copper pipe stationary. Find the magnitude and direction of the displacement of the pipe elbow when the water flow is turned on, raising the temperature of the pipe from 18.0°C to 40.2°C. (The coefficient of linear expansion of copper is

Answers

Answer:

The magnitude and direction are

7.638×10-4m

80.01°

Explanation:

We know that the coefficient of linear expansion for copper = 16.6×10^-6 m/m-C

ΔT = 40.2 - 18.0 = 28.5 C°

The expansion of horizontal pipe length can be calculated as

= (0.28)(16.6×10^-6)(28.5) = 13247×10^-8

= 0.0001325 m

The expansion of vertical pipe length = (1.28)(16.6×10^-6)(28.5) = 60557×10^-8 = 0.000752229 m

horizontal displacement = 0.1325 mm

= 1.356×10^-4m

vertical displacement = 0.75223mm

=7.5223×10-4m

size of total displacement can be calculated as

√(x²+y²)

Where x and y are vertical and horizontal displacement respectively

= √(0.1325)²+(0.75223)² =

= 0.7638 mm

= 7.638×10-4m

Angle below horizontal = arctan Θ

= 0.75223/0.1325

=5.6772

= arctan (5.6772)

= 80.01°

Therefore, the the magnitude and direction of the displacement of the pipe elbow when the water flow is turned at (7.638×10-4m) 0.7638 mm and 80.01°

If the magnetic field of an electromagnetic wave is in the +x-direction and the electric field of the wave is in the +y-direction, the wave is traveling in the

Answers

Answer:

The wave is travelling in the ±z-axis direction.

Explanation:

An electromagnetic wave has an oscillating magnetic and electric field. The electric and magnetic field both oscillate perpendicularly one to the other, and the wave travels perpendicularly to the direction of oscillation of the  electric and magnetic field.

In this case, if the magnetic field is in the +x-axis direction, and the electric field is in the +y-axis direction, we can say with all assurance that the wave will be travelling in the ±z-axis direction.

An ac generator consists of a coil with 40 turns of wire, each with an area of 0.06 m2 . The coil rotates in a uniform magnetic field B = 0.4 T at a constant frequency of 55 Hz. What is the maximum induced emf?

a. 625 V
b. 110 V
c. 421 V
d. 332 V
e. 200 V

Answers

Answer:

d. 332 V

Explanation:

Given;

number of turns in the wire, N = 40 turns

area of the coil, A = 0.06 m²

magnitude of the magnetic field, B = 0.4 T

frequency of the wave, f = 55 Hz

The maximum emf induced in the coil is given by;

E = NBAω

Where;

ω is angular velocity = 2πf

E = NBA(2πf)

E = 40 x 0.4 x 0.06 x (2 x π x 55)

E = 332 V

Therefore, the maximum induced emf in the coil is 332 V.

The correct option is "D"

d. 332 V

Water is draining from an inverted conical tank with base radius 8 m. If the water level goes down at 0.03 m/min, how fast is the water draining when the depth of the water is 6 m

Answers

Answer:

0.03/π m/min

Explanation:

See attached file pls

What is the wavelength of electromagnetic radiation which has a frequency of 3.818 x 10^14 Hz?

Answers

Answer:

7.86×10⁻⁷ m

Explanation:

Using,

v = λf.................. Equation 1

Where v = velocity of electromagnetic wave, λ = wave length, f = frequency.

make λ the subject of the equation

λ = v/f............... Equation 2

Note: All electromagnetic  wave have the same speed which is 3×10⁸ m/s.

Given: f = 3.818×10¹⁴ Hz

Constant: v = 3×10⁸ m/s

Substitute these values into equation 2

λ  =  3×10⁸/3.818×10¹⁴

λ  = 7.86×10⁻⁷ m

Hence the wavelength of the electromagnetic radiation is  7.86×10⁻⁷ m

The wavelength of this electromagnetic radiation is equal to [tex]7.86 \times 10^{-7} \;meters[/tex]

Given the following data:

Frequency = [tex]3.818\times 10^{14}\;Hz[/tex]

Scientific data:

Velocity of an electromagnetic radiation = [tex]3 \times 10^8\;m/s[/tex]

To determine the wavelength of this electromagnetic radiation:

Mathematically, the wavelength of an electromagnetic radiation is calculated by using the formula;

[tex]Wavelength = \frac{Speed }{frequency}[/tex]

Substituting the given parameters into the formula, we have;

[tex]Wavelength = \frac{3 \times 10^8}{3.818\times 10^{14}}[/tex]

Wavelength = [tex]7.86 \times 10^{-7} \;meters[/tex]

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Two automobiles are equipped with the same singlefrequency horn. When one is at rest and the other is moving toward the first at 20 m/s , the driver at rest hears a beat frequency of 9.0 Hz.

Requried:
What is the frequency the horns emit?

Answers

Answer: f ≈ 8.5Hz

Explanation: The phenomenon known as Doppler Shift is characterized as a change in frequency when one observer is stationary and the source emitting the frequency is moving or when both observer and source are moving.

For a source moving and a stationary observer, to determine the frequency:

[tex]f_{0} = f_{s}.\frac{c}{c-v_{s}}[/tex]

where:

[tex]f_{0}[/tex] is frequency of observer;

[tex]f_{s}[/tex] is frequency of source;

c is the constant speed of sound c = 340m/s;

[tex]v_{s}[/tex] is velocity of source;

Rearraging for frequency of source:

[tex]f_{0} = f_{s}.\frac{c}{c-v_{s}}[/tex]

[tex]f_{s} = f_{0}.\frac{c-v_{s}}{c}[/tex]

Replacing and calculating:

[tex]f_{s} = 9.(\frac{340-20}{340})[/tex]

[tex]f_{s} = 9.(0.9412)[/tex]

[tex]f_{s} =[/tex] 8.5

Frequency the horns emit is 8.5Hz.

A competitive diver leaves the diving board and falls toward the water with her body straight and rotating slowly. She pulls her arms and legs into a tight tuck position. What happens to her rotational kinetic energy

Answers

Answer: her rotational kinetic energy increases

A front wheel drive car starts from rest and accelerates to the right. Knowing that the tires do not slip on the road, what is the direction of the friction force the road applies to the front tires

Answers

Answer:

static friction acting opposite to the direction of travel

Explanation:

Because the Frictional force of the front wheels act to oppose the spinning, so, For the front wheels to roll without slipping, the friction must be static friction pointing in the direction of travel of the car.

Explanation:

During the new moon phase, why is the Moon not visible in the sky?

Answers

Answer:

Explanation:

The moon gets the light from the sun. When the moon lies between the sun and the earth, only the back portion of the moon gets the light from the sun. So the side facing the sun does not get any light and appears to be dark or does not appear at all.

Hope this helps

plz mark as brainliest!!!!!!!

Answer:

The moon is between the sun, and Earth and reflects light back towards the sun.

Explanation:

A P E X test answer. Just took the test and this is the correct answer.

A long straight solenoid has 800 turns. When the current in the solenoid is 2.90 amperes the average flux through each turn is 3.25×10−3Wb.
A. What is the inductance of the coil?
B. What must be the magnitude fo the rate of change of the current (di/dt) in order for the self-induced emf to equal 7.50 mV?

Answers

Answer:

Explanation:

Relation between flux and inductance is as follows

φ = Li

where φ is flux associated with induction of inductance L when a current i flows through it

putting the values

3.25 x 10⁻³ x 800 = L x 2.9

L = .9 H

for induced emf in an induction , the relation is

emf induced = L di / dt

Putting the values

7.5 x 10⁻³ = .9 x di / dt

di / dt = 8.33 x 10⁻³ A / s

(a) The self inductance of the solenoid is 0.897 H.

(b) The magnitude of the rate of change of the current is 0.00836 A/s.

The given parameters;

number of turns, N = 800 turnscurrent in the solenoid, I = 2.9 flux through the solenoid, Ф = 3.25 x 10⁻³ Wb

The self inductance of the solenoid is calculated as follows;

[tex]emf = \frac{d\phi}{dt}\\\\emf = \frac{Ldi}{dt} \\\\d\phi = Ldi\\\\\phi = BA\\\\NBA = LI\\\\L = \frac{NBA}{I} \\\\L = \frac{N\phi}{I} \\\\L = \frac{800 \times 3.25\times 10^{-3}}{2.9} \\\\L = 0.897 \ H\\\\[/tex]

The magnitude of the rate of change of the current is calculated as follows;

[tex]emf = L \frac{di}{dt} \\\\\frac{di}{dt} \ = \frac{emf}{L} \\\\\frac{di}{dt} = \frac{7.5 \times 10^{-3}}{0.897} \\\\\frac{di}{dt} = 0.00836 \ A/s[/tex]

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where c is the speed of light and G is the universal gravitational constant. RBH gives the radius of the event horizon of a black hole with mass ????. In other words, it gives the radius to which some amount of mass ???? would need to be compressed in order to form a black hole. The mass of the Sun is about 1.99×1030 kg. What would be the radius of a black hole with this mass?

Answers

Answer:

The radius of the black hole will be 2949.6 m.

Explanation:

The radius of this black hole will be the Schwarzschild radius of the mass of the sun

[tex]r_{s}[/tex] = [tex]\frac{2GM}{c^{2} }[/tex]

where

G is the gravitational constant = 6.67 x 10^-11 m^3⋅kg^-1⋅s^-2

M is the mass of the sun = 1.99×10^30 kg

c is the speed of light = 3 x 10^8 m/s

substituting values into the equation, we have

[tex]r_{s}[/tex] = [tex]\frac{2*6.67*10^{-11}*1.99*10^{30} }{(3*10^{8} )^{2} }[/tex] = 2949.6 m

The starter motor of a car engine draws a current of 140 A from the battery. The copper wire to the motor is 4.20 mm in diameter and 1.2 m long. The starter motor runs for 0.760 s until the car engine starts.Required:a. How much charge passes through the starter motor? b. How far does an electron travel along the wire while the starter motor is on?(mm)

Answers

Answer:

(a)106.4C

b)0.5676mm

Explanation:

(a)To get the charge that have passed through the starter then The current will be multiplied by the duration

I= current

t= time taken

Q= required charge

Q= I*t = 140*0.760 = 106.C

(b) b. How far does an electron travel along the wire while the starter motor is on?(mm)

diameter of the conductor is 4.20 mm

But Radius= diameter/2= 4.20/2=

The radius of the conductor is 2.1mm, then if we convert to metre for consistency same then

radius of the conductor is 0.0021m.

We can now calculate the area of the conductor which is

A = π*r^2

= π*(0.0021)^2 = 13.85*10^-6 m^2

We can proceed to calculate the current density below

J = 140/13.85*10^-6 = 10108303A/m

According to the listed reference:

Where e= 1.6*10^-19

n= 8.46*10^28

Vd = J/(n*e) = 10108303/ ( 8.46*10^28 * 1.6*10^-19 ) =0.0007468m/s=0 .7468 mm/s

Therefore , the distance traveled is:

x = v*t = 0.7468 * 0.760 = 0.5676mm

(a) The charge passes through the starter motor is 106.4C.

(b) An electron travel along the wire while the starter motor is on 0.5676mm.

Electron

Answer (a)

I= current

t= time taken

Q= required charge

Q= I*t

Q= 140*0.760

Q= 106.C

Answer (b)

The n electron travel along the wire while the starter motor is on:

Diameter of the conductor is 4.20 mm

Radius= diameter/2= 4.20/2

Radius =2.1mm

Radius of the conductor is 0.0021m.

A = π*r^2

A= π*(0.0021)^2

A= 13.85*10^-6 m^2

Where e= 1.6*10^-19

n= 8.46*10^28

Vd = J/(n*e) = 10108303/ ( 8.46*10^28 * 1.6*10^-19 )

Vd  =0.0007468m/s

Vd =0 .7468 mm/s

The distance traveled is:

x = v*t

x= 0.7468 * 0.760

x = 0.5676mm

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