Select four of the following that would increase the magnetic field of an electromagnet

Select Four Of The Following That Would Increase The Magnetic Field Of An Electromagnet

Answers

Answer 1

Answer:

The correct answers are: A, C, D, E

Explanation:

The magnetic field is a solenoid is given by

         B = μ₀  [tex]\frac{N}{L}[/tex]   I

where N is the number of turns, I the current and L length of the solenoid.

Using this equation let's examine the different responses to permute increasing the magnetic field

A) True. a thicker wire decreases the resistance and the current can increase the system.

B) False. If there is no voltage source there is no current in the system

C) True. the field is proportional to the number of turns

D) True. the magnetic moments of the core align with the field increasing its value

E) True. When the loops are closer together, more of them can fit per unit length

F) False. If the wire is shorter the number of turns decreases.

The correct answers are: A, C, D, E


Related Questions

A light year is the amount of time it takes for light from the Sun to reach the Earth.
True
False

Answers

true! a light year is the amount of time it takes for light from the sun to reach Earth!

Soap bubbles can display impressive colors, which are the result of the enhanced reflection of light of particular wavelengths from the bubbles' walls. For a soap solution with an index of refraction of 1.21, find the minimum wall thickness that will enhance the reflection of light of wavelength 711 nm in air.

Answers

Answer:

the minimum wall thickness that will enhance the reflection of light is 146.9 nm

Explanation:

Given the data in the question;

At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.

At the second interface, no shift occurs,

condition for constructive interference;

t = ( m + 1/2) × λ/2n

where m = 0, 1, 2, 3 . . . . . .

now, the condition for the constructive interference;

t = mλ/2n

where t is the thickness of the soap bubble,  λ is the wavelength of light and n is the refractive index of soap bubble.

so the minimum thickness of the film which will enhance reflection of light will be;

t[tex]_{min[/tex] =  ( m + 1/2) × λ/2n

we substitute

t[tex]_{min[/tex] =  ( 0 + 1/2) × 711 /2(1.21)

t[tex]_{min[/tex] = 0.5 × 711/2.42

t[tex]_{min[/tex] = 0.5 × 293.80165

t[tex]_{min[/tex] = 146.9 nm

Therefore,  the minimum wall thickness that will enhance the reflection of light is 146.9 nm

(5 Points)
a) At ground level, the pressure of the helium in a balloon is 1x105
Pa. The volume occupied by the helium is 9.6m The balloon is
released and it rises quickly through the atmosphere. Calculate
the pressure of the helium when it occupies a volume of 12m3.
(3 Marks)
b) A box is 15m below the surface of the sea. The density of sea-
water is 1020 kg/m.
Calculate the pressure on the box due to the sea-water.
(2 Marks)

Answers

Answer:

1. [tex]P_{2}[/tex] = 8 x [tex]10^{4}[/tex] Pa

2. P = 1.5 x [tex]10^{5}[/tex] N/[tex]m^{2}[/tex]

Explanation:

1. From Boyles' law;

[tex]P_{1}[/tex][tex]V_{1}[/tex] = [tex]P_{2}[/tex][tex]V_{2}[/tex]

[tex]P_{1}[/tex] = 1 x [tex]10^{5}[/tex] Pa

[tex]V_{1}[/tex] = 9.6 [tex]m^{3}[/tex]

[tex]V_{2}[/tex] = 12 [tex]m^{3}[/tex]

Thus,

1 x [tex]10^{5}[/tex] x 9.6 =  [tex]P_{2}[/tex] x 12

 [tex]P_{2}[/tex] = [tex]\frac{100000 x 9.6}{12}[/tex]

     = 80000

[tex]P_{2}[/tex] = 8 x [tex]10^{4}[/tex] Pa

2. Pressure, P = ρhg

where: ρ is the density of the fluid, h is the height/ depth and g is the acceleration due to gravity (9.8 m/[tex]s^{2}[/tex]).

Thus,

P = 1020 x 15 x 9.8

  = 149940

P = 1.5 x [tex]10^{5}[/tex] N/[tex]m^{2}[/tex]

Scientific work is currently under way to determine whether weak oscillating magnetic fields can affect human health. For example, one study found that driv- ers of trains had a higher incidence of blood cancer than other railway workers, possibly due to long expo- sure to mechanical devices in the train engine cab. Consider a magnetic field of magnitude 1.00 3 1023 T, oscillating sinusoidally at 60.0 Hz. If the diameter of a red blood cell is 8.00 mm, determine the maximum emf that can be generated around the perimeter of a cell in this field.

Answers

Answer:

fem = 7.58 10⁻⁵ V

Explanation:

For this exercise we use Faraday's law

          fem =   [tex]- \frac{d \Phi _B}{dt}[/tex]

the magnetic flux is

          Ф_B = B. A = B A cos θ

Tje bold are vectros.  Suppose the case where the normal to the surface of the red blood cell is parallel to the field, therefore the angle is zero and the cos 0 = 1

The red blood cell area is

          A =π r²

       

indicate that the diameter is r = 8.00 mm = 8.00 10⁻³ m

the magnetic field has a frequency of f=60 Hz, and B₀ = 1.00 10⁻³T,  therefore we can write it

          B = B₀ sin (wt) = B₀ sin( 2π f t)

           

we substitute

           fem = - A dB / dt

           fem = - A B₀  [tex]\frac{ d (sin ( 2\pi f t)}{dt}[/tex]

           fem = - π r² Bo (2πf  cos 2πft)

the maximum electromotive force occurs when the function is ±1

       

           fem = 2 π² r² B₀ f

let's calculate

           fem = 2π²  (8.00 10⁻³)²  1.00 10⁻³ 60

           fem = 7.58 10⁻⁵ V

A positive charge Q2 is uniformly distributed over a nonconducting disc of radius a which has a concentric circular hole of radius b. At the center of the hole there is another nonconducting disc of radius d where a charge Q1 is uniformly distributed.
a) Find the surface charge density of the disc with the hole σ2.
b) Find the surface charge density 01 of the disc of radius d.
c) Find the total charge enclosed by the circle of radius

Answers

Answer:

a)    σ = [tex]\frac{Q_1}{ a^2 - b^2}[/tex] ,  b)  σ = [tex]\frac{Q_2}{d^2}[/tex] , c) Q_ {total} = Q₁ + Q₂,  σ_ {net} = [tex]\frac{Q_1 + Q_2}{\pi \ a^2}[/tex]

Explanation:

a) The very useful concept of charge density is defined by

          σ = Q / A

In this case we have a circular disk

The are of a circle is

         A = π r²

in this case we have a hole in the center of radius r = b, so

         A_net = π r² - π r_ {hollow} ²

         A_ {net} = π (a² - b²)

whereby the density is

          σ = [tex]\frac{Q_1}{ a^2 - b^2}[/tex]

b) The density of the other disk is

          σ = Q₂ / A₂

          σ = [tex]\frac{Q_2}{d^2}[/tex]

c) The total waxed load is requested by the larger circle

           Q_ {total} = Q₁ + Q₂

the net charge density, in the whole system is

          σ = [tex]\frac{Q_{total} }{ A_{total} }[/tex]

the area  is

          A_{total} = π a²

since the other circle is inside, we are ignoring the space between the two circles

          σ_ {net} = [tex]\frac{Q_1 + Q_2}{\pi \ a^2}[/tex]

If you could help me please.
1) Does a 1 kg object weight 9.8 newtons on the moon? why?
2) How much does a 3-kg object weigh (on earth) in newtons?
3) How much does a 20-kg object weigh (on earth) in newton?
4) What must happen for the mass of an object to change?
5) What are 2 ways the weight of an object can change?

Answers

1) Does a 1 kg object weight 9.8 newtons on the moon? why?

No. 1kg of mass does not weigh 9.8N on the moon.

Weight = (mass) x (gravity).

Gravity is 9.8 m/s² on Earth, but gravity is only 1.62 m/s² on the moon.

2) How much does a 3-kg object weigh (on earth) in newtons?

Weight = (mass) x (gravity)

Gravity = 9.8 m/s² on Earth.

Weight = (3 kg) x (9.8 m/s² )

Weight = 29.4 N

3) How much does a 20-kg object weigh (on earth) in newton?

Weight = (mass) x (gravity)

Gravity = 9.8 m/s² on Earth.

Weight = (20 kg) x (9.8 m/s² )

Weight = 196 N

4) What must happen for the mass of an object to change?

When an object moves, its mass increases.  The faster it moves, the greater its mass gets.  But this is all part of Einstein's "Relativity".  The object has to move at a significant fraction of the speed of light before any change can be noticed or measured.  So as far as we are concerned, in everyday life, the mass of an object doesn't change, no matter where it is, or what you do to it.

5) What are 2 ways the weight of an object can change?

First, remember that the mass of an object doesn't change, no matter where it is, what you do to it, or what else is around it.

But its weight can change, because its weight depends on the strength of gravity in the place where the object is, and that gravity is the result of what else is around it in the neighborhood.  So the weight can change even though the mass doesn't.

The weight of an object changes if you take it to a place where gravity is stronger or weaker.

Let's say we have an object whose mass is 90.72 kilograms.  Like me !    

As long as I stay on earth, where gravity is 9.8 m/s² , I weigh 889 Newtons  (200 pounds).

. . . Fly me to the moon. Gravity = 1.62 m/s²  Weight = 147 Newtons (33 lbs)

. . . Drag me to Jupiter.  Gravity = 24.8 m/s²  Weight = 2,249 N (506 pounds)

My mass never changed, but my weight sure did.

An atom of tin has an atomic number of 50 and a mass number of 119. How many protons, electrons, and neutrons are found in one neutral atom of tin?
O 50 protons, 69 electrons, 50 neutrons
O 50 protons, 50 electrons, 69 neutrons
69 protons, 50 electrons, 69 neutrons
69 protons, 69 electrons, 50 neutrons

Answers

Answer:

50 protons 50 electrons and 69 neutrons...

Explanation:

the number of protons is equal to number of electrons. then mass number is the sum of protons and neutrons in a nucleus so for we to get the number of neutrons we take the mass number subtract the protons number.

a cohesive force between the liquids molecules is responsible for the fluids is called​

Answers

Answer:

static force

Explanation:

mark me brainliest

A hair dryer draws a current of 12.8 A.

(a)How many minutes does it take for
6.8 x 10° C of charge to pass through the
hair dryer? The fundamental charge is
1.602 x 10-19 C.
Answer in units of min.

(b)How many electrons does this amount of
charge represent?
Answer in units of electrons.

Answers

Answer:

(a) 8.85×10⁻³ minutes

(b) 4.24×10¹⁹ electrons

Explanation:

(a) Using,

Q = it............................. Equation 1

Where Q = quantity of charge, i = current, t = time.

Make t the subject of the equation

t = Q/i............................. Equation 2

Given: Q = 6.8×10⁰ C, i = 12.8 A

Substitute these values into equation 2

t = 6.8×10⁰/12.8

t = 8.85×10⁻³ minutes

(b) n = Q/(1.602×10⁻¹⁹)................. Equation 3

Where n = number of electrons.

Given: Q = 6.8×10⁰ C

Substitute into equation 2

n = 6.8×10⁰/1.602×10⁻¹⁹

n = 4.24×10¹⁹ electrons

(a) The time taken by the charge to flow from the drier will be [tex]\d8.85[/tex]×[tex]\d10^{-3}[/tex]minutes

(b) Amount of the electrons in the charge will be  [tex]\d4.24[/tex]×[tex]\d10^{19}[/tex] electrons

What will be the time of the charge and number of the electrons in the charge ?

As we know Q = IT

Where Q = quantity of charge, i = current, T = time.

From the above equation

                    T= Q/I.

Given: Q = [tex]6.8\times\d10^{0}[/tex] C, i = 12.8 A

Substitute these values  

T=  [tex]6.8[/tex]×[tex]\d10^{0}[/tex] /12.8

T =  [tex]8.85[/tex]×[tex]\d10^{-3}[/tex] minutes

Now the number of the electrons present in the charge will be

n = Q/( [tex]1.602[/tex]×[tex]\d10^{-19}[/tex])

Where n = number of electrons.

Given: Q = [tex]6.8\times\d10^{0}[/tex] C

Substitute Value of Q  

n =  [tex]6.8\times\d10^{0}[/tex]/ [tex]1.602\times\d10^{-19}[/tex]

n = [tex]4.24\times\d10^{19}[/tex] electrons

Thus

(a)The time taken by the charge to flow from the drier will be [tex]\d8.85[/tex]×[tex]\d10^{-3}[/tex]minutes

(b)Amount of the electrons in the charge will be  [tex]\d4.24[/tex]×[tex]\d10^{19}[/tex] electrons

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The amount of light that enters the pupil is controlled by the:
retina.
lens.
inis.

Answers

Answer: The amount of light that enters the pupil is controlled by the Iris

Explanation:

The gravitational force between two objects with masses 1kg and 28kg separated by a distance 7m is ____________10-11 N.


a.
3.81

b.
26.68

c.
9151.24

d.
1307.32
Hhhhhellllppp fastt​

Answers

Answer:

a. 3.81

Explanation:

F = GMm/r^2

F = (6.67 x 10^-11 x 28 x 1) / 7^2

F = 3.81 x 10^-11 N

why food cook faster with salt water than cook with pure water​

Answers

Answer:

oil heats faster

Explanation:

When salt is added, it makes it harder for the water molecules to escape from the pot and enter the gas phase, which happens when water boils

The average mean distance of Saturn from the sun is

Answers

Answer:

From an average distance of 886 million miles (1.4 billion kilometers), Saturn is 9.5 astronomical units away from the Sun. One astronomical unit (abbreviated as AU), is the distance from the Sun to Earth. From this distance, it takes sunlight 80 minutes to travel from the Sun to Saturn.

we have that from the Question"The average mean distance of Saturn from the sun is" it can be said that  Tthe average mean distance of Saturn from the sun is

A distance of 1427 x 10^6 km or 886 696 691 miles

From the Question we are told

The average mean distance of Saturn from the sun is

Generally

The Sun is the star of the milky way galaxy and its distance from every planet in the milky way determines in one way or another its properties and in-habitability

Saturn being a Planet of the milky way we see that Saturn is a significant distance away from sun

A distance of 1427 x 10^6 km or 886 696 691 miles

Therefore

The average mean distance of Saturn from the sun is

A distance of 1427 x 10^6 km or 886 696 691 miles

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What are some possible factors that can be the X
and Y axis of a motion graph?

Answers

Answer:

x-Speed/velocity

y-time.

Explanation:

because Speed is a rate of change of distance while time how long it takes a a car to move to a specific point

an iron Tyre of diameter 50cm at 288k is to be shrank on to a wheel of diameter 50.35cm.To what temperature must the tyre be heated so that it will slip over the wheel with a radial gap of 0.5mm.Linear expansivity of iron is 0.000012k-1

Answers

Answer:

The answer should be D

Explanation:

Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine:

Answers

This question is incomplete, the complete question is;

Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine: the angular velocity of Bxy rotating frame (ω).

Answer:

the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s

Explanation:

Given the data in the question and image below and as illustrated in the second image;

distance S = 40 m

V[tex]_B[/tex] = 54 km/hr

V[tex]_A[/tex] = 72 km/hr

α = 100 m

now, angular velocity of Bxy will be;

ω[tex]_B[/tex] = V[tex]_B[/tex] / α

so, we substitute

ω[tex]_B[/tex] = ( 54 × 1000/3600) / 100

ω[tex]_B[/tex] = 15 / 100

ω[tex]_B[/tex] = 0.15 rad/s

Therefore, the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s

12. Identify the Leader​

Answers

^^^^ nelson mandela is who it is

23
In order for a 12 Volt power source
to produce a current of 0.085 amps,
a resistance of...
[?] Ohms is needed.
Enter
Haven't learned this yet.

Answers

Answer:

141.18 ohms

Explanation:

From the question given above, the following data were obtained:

Voltage (V) = 12

Current (I) = 0.085 A

Resistance (R) =?

The resistance needed can be obtained as follow:

V = IR

12 = 0.085 × R

Divide both side by 0.085

R = 12 / 0.085

R = 141.18 ohms

Therefore, a resistor of resistance 141.18 ohms is needed.

On the Moon's surface, lunar astronauts placed a corner reflector, off which a laser beam is periodically reflected. The distance to the Moon is calculated from the round-trip time. The Earth's atmosphere slows down light. Assume the distance to the Moon is precisely 3.84×108 m, and Earth's atmosphere (which varies in density with altitude) is equivalent to a layer 30.0 km thick with a constant index of refraction n=1.000293. What is the difference in travel time for light that travels only through space to the moon and back and light that travels through the atmosphere and space?

Answers

Answer:

a) space only    t = 1.28 s

b) space+ atmosphere   t_ {total} = 1.28000003 s

Explanation:

The speed of light in each material medium is constant, which is why we can use the uniform motion relations

           v= x / t

           

a) let's look for time when it only travels through space

          t = x / c

          t = 3.84 10⁸/3 10⁸

          t = 1.28 s

b) we look for time when it travels part in space and part in the atmosphere

space

as it indicates that the atmosphere has a thickness of e = 30 10³ m

           t₁ = (D-e) / c

           t₁ = (3.84 10⁸ - 30.0 10³) / 3 10⁸

           t₁ = 1.2799 s

           

atmosphere

             

we use the refractive index

           n = c / v

           v = c / n

we substitute in the equation of time

           t₂ = e n / c

           t₂ = 30 10³   1,000293 /3 10⁸

           t₂ = 1.000293 10⁻⁴ s

therefore the total travel time is

           t_ {total} = t₁ + t₂

           t_ {total} = 1.2799+ 1.000293 10⁻⁴

           t_ {total} = 1.28000003 s

we can see that the time increase due to the atmosphere is very small

PLEASE CLICK ON THIS IMAGE I NEED HELP

Answers

Answer:

Explanation:

you can say the law of superpoition can tell us that each rock layer is older than the one above it. So, the relative age of the rock or fossil in the rock or fossil in the rock is older if it is farther down in the rock layers. hope it helps

Which of the following changes when an unbalanced force acts on an object?
A. mass
B. motion
C. inertia
D. weight

Answers

The answer is Motion

If you blow across the open end of a soda bottle and produce a tone of 250 Hz, what will be the frequency of the next harmonic heard if you blow much harder?

___Hz

Answers

Answer:

Generally, the lowest overtone for a pipe open at one end and closed would be at  y / 4  where y represents lambda, the wavelength.

Since F (frequency) = c / y       Speed/wavelength

F2 / F1 = y1 / y2      because c is the same in both cases

F2 = y1/y2 * F1

F2 = 3 F1 = 750 /sec

Note that L = y1 / 4 = 3 y2 / 4 for these wavelengths to fit in the pipe

and y1 = 3 y2

The second harmonic will be three times the first harmonic. The answer is 750 Hz

VIBRATION OF WAVES IN PIPES

Closed pipes have odd multiples of frequencies or harmonics. That is,

If  [tex]F_{0}[/tex] = fundamental frequency = first harmonic

[tex]F_{1}[/tex] = 3[tex]F_{0}[/tex] = second harmonic

[tex]F_{2}[/tex] = 5[tex]F_{0}[/tex] = third harmonic

[tex]F_{3}[/tex] = 7[tex]F_{0}[/tex] = fourth harmonic

Let assume that the first harmonic is 250 Hz, If you blow it much harder, second, third or fourth harmonic can be produced.

By using the formula above,

second harmonic will be 3 x 250 = 750Hz

Therefore, the frequency of the next harmonic heard if you blow much harder will be 750 Hz

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A cat runs at 2 m/s for 3 s and then slows to a stop with an acceleration of 0.80 m/s2. What is
the cat's displacement during this motion?
A. 6m
B. 2.5 m
C. 8.5 m
D. 4.5 m
20
А
B
D​

Answers

Answer:

see that the correct one is B

Explanation:

To solve this exercise let us use the kinematic relations

           v² = v₀² - 2 a x

as they indicate that the car stops, therefore the final speed is yield v = 0

          x = v₀² / 2a

let's calculate

          x = 2²/(2 0.8)

         x = 2.5 m / s²

When reviewing the answers we see that the correct one is B

a boy throws a ball straight up into the air it reaches the highest point of its flight after 4 seconds how fast was the ball going when it left the boy's hand

Answers

Answer:

Gravity pulls down on the ball at g=-9.81 m/s^2. Up is positive, down is negative. The ball started at a certain initial velocity of Vi m/s. Time it took is t=4s. Final velocity is Vf=0 m/s, because at the highest point the ball stops moving.

In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the first light source has a wavelength of 640 nm. Two different interference patterns are observed. If the 10th order bright fringe from the first light source coincides with the 12th order bright fringe from the second light source, what is the wavelength of the light coming from the second monochromatic light source

Answers

Answer:

533.33 nm

Explanation:

Since dsinθ = mλ  for each slit, where m = order of slit and λ = wavelength of light. Let m' = 10 th order fringe of the first slit of wavelength of light, λ = 640 nm and m"= 12 th order fringe of the second slight of wavelength of light, λ'.

Since the fringes coincide,

m'λ = m"λ'

λ' = m'λ/m"

= 10 × 640 nm/12

= 6400 nm/12

= 533.33 nm

The speed limmit on an interstate highway is posted at 75mi/h. What is the speed in kilometers per hour? In feet per second?

Answers

I uploaded the answer to a file hosting. Here's link:

tinyurl.com/wpazsebu


1. Define force and give its Sl unit​

Answers

Answer:

Force is an external agency that changes or tends to change the state of body from rest to motion or motion to rest.

The SI unit of force is newton(N)

starting from rest, your bicycle can reach a speed of 4.0 m/s in 50 s. Assuming that your bicycle accelerates at a constant rate, what is its acceleration?

Answers

Answer:

0.08 ms^-2

Explanation:

by using v= u + at

initial velocity is zero as it is starting from rest

4= 0 + a x 50

4/50 = a = 0.08 ms^-2

At position A within a tube containing fluid that is moving with steady laminar flow, the speed of the fluid is 12.0 m/s and the tube has a diameter 12.00 cm. At position B, the speed of the fluid is 18.0 m/s and the tube has a diameter 6.00 cm. What is the ratio of the density of the fluid at position A to the density of the fluid at position B

Answers

Answer:

0.375

Explanation:

For incompressible flow, we know that;

ρ1•v1•A1 = ρ2•v2•A2

Where;

ρ1 = density of fluid at position A

v1 = speed of fluid at position A

A1 = area of tube

ρ2 = density of fluid at position B

v2 = speed of fluid at position B

A2 = area of tube

We want to find ratio of the density of the fluid at position A to the density of the fluid at position B.

Thus;

ρ1/ρ2 = (v2•A2)/(v1•A1)

Now, the tube will have the same height.

But we are given;

diameter of A = 12.00 cm = 0.12 m

diameter of B = 6 cm = 0.06 m

Thus;

A1 = π(d²/4)h = πh(0.12²/4)

A2 = πh(0.06²/4)

We are also given;

v1 = 12 m/s

v2 = 18 m/s

Thus;

ρ1/ρ2 = (18 × πh(0.06²/4))/(12 × πh(0.12²/4))

πh/4 will cancel out to give;

ρ1/ρ2 = (18 × 0.06²)/(12 × 0.12²)

ρ1/ρ2 = 0.375

Two Carnot heat engines are operating in series such that the heat sink of the first engine serves as the heat source of the second on. If the source temperature of the first engine is 1300 K and the sink temperature of the second engine is 300 K and the thermal efficiencies of both engines are the same, the temperature of the intermediate reservoir is

Answers

Answer:

the temperature of the intermediate reservoir is 624.5 K

Explanation:

Given the data in the question  

The two Carnot heat engines are operating in series;

[ T[tex]_H[/tex] ]

  ↓

((1)) ⇒ W[tex]_{out[/tex]

  ↓

[ T[tex]_M[/tex] ]

   ↓

 ((2)) ⇒ W[tex]_{out[/tex]

[ T[tex]_L[/tex] ]

The maximum possible efficiency for any heat engine is the Carnot efficiency;

η[tex]_{rev[/tex] = 1 - [tex]\frac{T_L}{T_H}[/tex]

the thermal efficiencies if both engines are the same will be;

η[tex]_A[/tex] = η[tex]_B[/tex]

1 -  [tex]\frac{T_M}{T_H}[/tex] = 1 - [tex]\frac{T_L}{T_M}[/tex]

1 - 1 -  [tex]\frac{T_M}{T_H}[/tex] = - [tex]\frac{T_L}{T_M}[/tex]

-  [tex]\frac{T_M}{T_H}[/tex] = - [tex]\frac{T_L}{T_M}[/tex]

[tex]\frac{T_M}{T_H}[/tex] =  [tex]\frac{T_L}{T_M}[/tex]

T[tex]_M[/tex]² = T[tex]_L[/tex] × T[tex]_H[/tex]

T[tex]_M[/tex] = √(T[tex]_L[/tex] × T[tex]_H[/tex])

source temperature of the first engine T[tex]_H[/tex] = 1300 K

sink temperature of the second engine T[tex]_L[/tex] = 300 K

we substitute

T[tex]_M[/tex] = √(300 × 1300)

T[tex]_M[/tex] = √390000

T[tex]_M[/tex] = 624.4998 K ≈ 624.5 K

Therefore, the temperature of the intermediate reservoir is 624.5 K

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