Example Problem
The potential energy of an object is given by U(x) = 8x2 - x4, where U is in joules and x is in
(a) Determine the force acting on this object.
(b) At what positions is this object in equilibrium?
(c) Which of these equilibrium positions are stable and which are unstable?
metres.
111 Unit 2 Concepts and Definitions Prof Mark Lester

Exam Part B Example
A neutron of mass m moving with velocity v collides head-on and elastically with a stationary nucleus of mass M.
(a) Show that the velocity of the nucleus after the collision, U, is given by
U= 2m v (m+M)
(b) Hence show that the neutron loses a fraction f of its energy where
f= 4mM (m+M)
10marks 5 marks
(c) A fast neutron enters a target of carbon nuclei which may be assumed to have masses 12 times that of the neutron. How many head-on collisions will it take
before the neutron loses 95% of its energy?
4 marks
(d) Suggest one reason why in a real reactor a neutron is likely to make more
collisions with the moderator nuclei before losing this much energy
2
1 mark

Answers

Answer 1

Answer:

Part A

a)  F = -16x + 4,  b)  x = 0.25 m, c) STABLE

Explanation:

Part A

a) Potential energy and force are related

          F = [tex]- \frac{dU}{dx}[/tex]- dU / dx

          F = - (8 2x -4)

          F = -16x + 4

b) The object is in equilibrium when the forces are zero

          0 = -16x + 4

          x = 4/16

          x = 0.25 m

c) An equilibrium position is called stable if with a small change in position, the forces make it return to the initial position, in case the forces make it move away it is called unstable.

In this case there is only one equilibrium point

by changing the position a bit

           x ’= x + Δx

we substitute

          F ’= - 16 x’ + 4

          F ’= - 16 (x + Δx) + 4

          F ’= (-16x +4) - 16 Δx

at equilibrium position F = 0

          F ’= 0 - 16 Δx

we can see that the body returns to the equilibrium position, therefore it is STABLE

PART B

This is an exercise in body collisions, let's define the system formed by the two bodies in such a way that the forces during the collisions are internal and the moment is conserved

initial instant. Before the shock

        p₀ = m v

final instant. After the crash

        p_f = (m + M) v_f

We have two possibilities: an elastic collision in which the bodies separate, each one maintaining its plus, and an INELASTIC collision where the neutron is absorbed by the nucleus and the final mass is M '= m + M, in this case they indicate that the collision is elastic

          p₀ = pf

          mv = mv ’+ M v_f

in the case of the elastic collision, the kinetic energy is conserved

        K₀ = K_f

        ½ m v² = ½ m v’² + ½ M v_f²

we write the system of equations

        mv = mv ’+ M v_f (1)

         m (v² -v'²) = M v_f ²

         

         m (v - v ’) = M v_f

         m (v-v ’) (v + v’) = M v_f

         

        v + v ’= v_f

we substitute in equation 1 and solve

         v ’=[tex]\frac{m -M }{m+M } \ vo[/tex]

         v_f = [tex]\frac{2m}{m+M} \ v_o[/tex]

the mechanical energy of the neutron is

  initial

          Em₀ = K = ½ m v²

final moment

          Em_f = K + U = ½ m v_f ² + U

U is the energy lost in the collision

total energy is conserved

          Em₀ = Em_f

          ½ m v² = ½ m v_f ² + U

         U = ½ m (v² -v_f ²)

         U = ½ m [v² - ( [tex]\frac{m-M}{m+M}[/tex]  v)² ]

 

       U = ½ m v² [1- ( [tex]\frac{m-M}{m+M}[/tex] )² ]

       U = ½ m v2 [ [tex]\frac{2M}{m+M}[/tex]]

       

       U = [tex]\frac{2 mM}{m +M } \ v^2[/tex]

Let's do the same calculations for the nucleus

initial     Em₀ = 0

final        Em_f = K + U = ½ M v_f ² + U

            Em₀ = Em_f

            0 = K + U

            U = -K

            U = - ½ M v_f ²

            U = - ½ M [ [tex]\frac{2m}{m+M} \ v[/tex] ]²

            U = [tex]\frac{2 m M }{m+M} \ v^2[/tex]  

We can see that we obtain the same result, that is, the potential energy lost by the neutron is equal to the potential energy gained by the nucleus.

 

b) the fraction of energy lost

          f = U / Em₀

          f = 4 m M / m + M        

c) let's calculate the fraction of energy lost in a collision

          m = 1.67 10⁻²⁷ kg

          M = 12 1.67 10⁻²⁷= 20 10⁻²⁷ kg

         f = 4 1.6 20 / (1.6+ 20)    10⁻²⁷

         f = 5.92 10⁻²⁷ J

the energy of a fast neutron is greater than 1 eV

         Eo = 1 eV (1.67 10⁻¹⁹ J / 1eV) = 1.67 10⁻¹⁹ J

Let's use a direct portion rule if in a collision f loses in how many collisions it loses 0.95Eo

         #_collisions = 0.95 Eo / f

         #_collisions = 0.95 1.67 10⁻¹⁹ / 5.92 10⁻²⁷

         #_collisions = 2.7 10⁷ collisions


Related Questions

Which of the globes in Figure 7.9 will light up?

Answers

I believe it would be A because that’s the only one that seems to complete the circuit.

pls someone help me pls…. and pls explain to me how

Answers

Answer:

1.12 × 10⁴ m/s

Explanation:

The escape velocity of the object v = √(2GM/R) where G = gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², M = mass of the Earth = 6 × 10²⁴ kg and R = radius of the Earth = 6.4 × 10⁶ m

Since v = √(2GM/R)

Substituting the values of the variables into the equation, we have

v = √(2GM/R)

v = √(2 × 6.67 × 10⁻¹¹ Nm²/kg² × 6 × 10²⁴ kg/6.4 × 10⁶ m)

v = √(13.34 × 10⁻¹¹ Nm²/kg² × 6 × 10²⁴ kg/6.4 × 10⁶ m)

v = √(80.04 × 10⁻¹¹ × 10²⁴Nm²/kg/6.4 × 10⁶ m)

v = √(80.04 × 10¹³Nm²/kg ÷ 6.4 × 10⁶ m)

v = √(80.04 ÷ 6.4 × 10¹³ ÷ 10⁶Nm/kg)

v = √(12.50625 × 10⁷ Nm/kg)

v = √(125.0625 × 10⁶ Nm/kg)

v = 11.18 × 10³ m/s

v = 1.118 × 10 × 10³ m/s

v = 1.118 × 10⁴ m/s

v ≅ 1.12 × 10⁴ m/s

A 250g object hangs from a spring that has a spring constant of 48.0 N/m and oscillates with an amplitude of 5.42cm

1)The magnitude of the objects acceleration when the displacement is 4.27 cm (down) is __ m/s^2

2)Given that the object has an amplitude of 5.42 cm the maximum speed that the object is __m/s

Answers

Explanation:

Given that,

Mass of an object, m = 250 g = 0.25 kg

Spring constant, k = 48 N/m

The amplitude of the oscillation, A = 5.42 cm = 0.0542 m

1. At equilibrium,

ma = kx

Where

a is the acceleration of the object

So,

[tex]a=\dfrac{kx}{m}\\\\a=\dfrac{48\times 0.0542}{0.25}\\\\a=10.4\ m/s^2[/tex]

2. The maximum speed of the object is :

[tex]v=A\omega\\\\v=A\sqrt{\dfrac{k}{m}}\\\\v=0.0542\times \sqrt{\dfrac{48}{0.25}}\\\\v=0.75\ m/s[/tex]

Hence, this is the required solution.

A tennis player hits a ball hard and 0.80 seconds later hears the echo from a wall. The speed of sound in air is 340m/s how far away is the player from the wall

Answers

Answer:

136 m.

Explanation:

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

Time (t) = 0.80 s

Speed of sound (v) = 340 m/s

Distance (x) =?

The distance of the player from the wall can be obtained as follow:

v = 2x / t

340 = 2x / 0.80

Cross multiply

2x = 340 × 0.80

2x = 272

Divide both side by 2

x = 272 / 2

x = 136 m

Thus, the distance of the player from the wall is 136 m

What's an area of the earth that suggest that " the Earth is like one giant jigsaw" idea could be true.

Answers

An area of the earth that suggests the earth is one big jigsaw is, the outer surface.


4 A student says that he has made a magnetic field with some iron filings. What would
you say to him?

Answers

Answer:

Yes but no

Explanation:

Here the kid cant say that he has created s magnetic field as iron fillings do not have magnetic properties

A tray containing 0.20kg of water at 20degree celsius is placed in a freezer. The temperature of the water drops to 0degree celsius in 10 minutes. Calculate


a) The energy lost by the water asit cools to 0 degree celsius.

b) The average rate at which the water is losing energy in J/s.

c) Estimate the time taken for the water at 0 degree celsius to turn completely into ice.

d) state any assumptios you make​

Answers

Answer:

a. Energy lost, Q = 16,800 Joules.

b. Power = 28 J/s

c. Time, t = 2357.14 seconds

d. I assumed that the ice remained at a temperature of zero degrees Celsius (0°C). Also, I assumed that the heat is being lost at a constant rate.

Explanation:

Given the following data;

Mass = 0.20 kgInitial temperature, T1 = 20°CFinal temperature = 0°CTime = 10 minutes

a. To find the energy lost by the water as it cools to 0 degree celsius;

Mathematically, heat capacity is given by the formula;

[tex] Q = mcdt [/tex]

Where;

Q represents the heat capacity or quantity of heat.M represents the mass of an object.C represents the specific heat capacity of water.dt represents the change in temperature.

dt = T2 - T1

dt = 20 - 0

dt = 20°C

We know that the specific heat capacity of water is equal to  4200 J/kg°C

Substituting the values into the formula, we have;

[tex] Q = 0.20 * 4200 * 20 [/tex]

Energy lost, Q = 16,800 Joules.

b. To find the average rate at which the water is losing energy in J/s by using the following formula;

[tex] Power = \frac {energy}{time} [/tex]

First of all, we would have to convert the value of time in minutes to seconds.

Conversion:

1 minute = 60 seconds

10 minutes = X seconds

Cross-multiplying, we have;

X = 60 * 10

X = 600 seconds

Substituting the values into the formula, we have;

[tex] Power = \frac {16800}{600} [/tex]

Power = 28 J/s

c. To estimate the time taken for the water at 0 degree celsius to turn completely into ice;

We know that the latent heat of fusion of water is equal to 3.3 * 10⁵ J/kg.

Mathematically, the latent heat of fusion is calculated by using the formula;

Energy, Q = ml = pt

Substituting the values into the formula, we have;

0.20 * 3.3 * 10⁵ = 28 * t

0.20 * 330000 = 28t

66000 = 28t

[tex] t = \frac {66000}{28} [/tex]

Time, t = 2357.14 seconds.

d. The assumption made is that, the ice remained at a temperature of zero degrees Celsius (0°C). Also, I assumed that the heat is being lost at a constant rate.

Multi-part question If a galaxy moving away from the Earth has a speed of 1000 km/s and emits 656 nm light characteristic of hydrogen (the most common element in the universe). What wavelength would we observe on the Earth

Answers

Answer:

658.2 nm

Explanation:

Since the galaxy is moving at relavitistic speed, we use the equation for relativistic Doppler shift of light.

So, the wavelength of light observed on the Earth is λ

λ = λ'([tex]\sqrt{\frac{ 1 + \frac{v}{c} }{1 - \frac{v}{c} } }[/tex])

where  λ' = wavelength of light emitted by galaxy = 656 nm, v = speed of galaxy = 1000 km/s (positive since the galaxy is moving away from the Earth) and c = speed of light = 300000 km/s

So, substituting the values of the variables into the equation, we have

λ = λ'(√[{1 + (v/c)}/(1 - (v/c)]

λ = 656 nm(√[{1 + (1000 km/s/300000 km/s)}/(1 - (1000 km/s/300000 km/s)]

λ = 656 nm(√[{1 + 1/300}/(1 - 1/300]

λ = 656 nm(√[{(300 + 1)/300}/{(300 - 1)/300}]

λ = 656 nm(√[{(301)/300}/{(299)/300}]

λ = 656 nm(√[301/299])

λ = 656 nm(√1.0067)

λ = 656 nm × 1.0033

λ = 658.19 nm

λ ≅ 658.2 nm

So, the wavelength observed on Earth is 658.2 nm

Thiết bị nào sau đây không phải là nguồn điện
A. Ắc quy
B. Pin
C. Máy phát điện
D. Bóng đèn điện

Answers

shsbbsnsjsbsnsnsnsnsnsndndndnd

Two manned satellites approach one another at a relative velocity of v=0.190 m/s, intending to dock. The first has a mass of m1=4.00×103 kg and the second a mass of m2=7.50×103 kg. If the two satellites collide elastically rather than dock, what is their final relative velocity?

Answers

Answer:

Their final relative velocity is 0.190 m/s

Explanation:

The relative velocity of the satellites, v = 0.190 m/s

The mass of the first satellite, m₁ = 4.00 × 10³ kg

The mass of the second satellite, m₂ = 7.50 × 10³ kg

Given that the satellites have elastic collision, we have;

[tex]v_2 = \dfrac{2 \cdot m_1}{m_1 + m_2} \cdot u_1 - \dfrac{m_1 - m_2}{m_1 + m_2} \cdot u_2[/tex]

[tex]v_2 = \dfrac{ m_1 - m_2}{m_1 + m_2} \cdot u_1 + \dfrac{2 \cdot m_2}{m_1 + m_2} \cdot u_2[/tex]

Given that the initial velocities are equal in magnitude, we have;

u₁ = u₂ = v/2

u₁ = u₂ = 0.190 m/s/2 = 0.095 m/s

v₁ and v₂ = The final velocities of the satellites

We get;

[tex]v_1 = \dfrac{2 \times 4.0 \times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 - \dfrac{4.0 \times 10^3- 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095[/tex]

[tex]v_2 = \dfrac{ 4.0 \times 10^3 - 7.50\times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 + \dfrac{2 \times 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095[/tex]

The final relative velocity of the satellite, [tex]v_f[/tex] = v₁ + v₂

∴ [tex]v_f[/tex] = 0.095 + 0.095 = 0.190

The final relative velocity of the satellite, [tex]v_f[/tex] = 0.190 m/s

3 write the three laws given by kepler.How did they help Newton to arrive at the inverse square law of gravity?​

Answers

Answer:

Kepler's laws apply: First Law: Planetary orbits are elliptical with the sun at a focus. Second Law: The radius vector from the sun to a planet sweeps equal areas in equal times. Third Law: The ratio of the square of the period of revolution and the cube of the ellipse semimajor axis is the same for all planets.

An airplane flying at 116 m/s. E, is accelerated uniformly at the rate of 9.2 m/s2, E, for 13 s. What is its final velocity in m/s?

Answers

Answer:

235.6m/s

Explanation:

you have to use one of the kinematic formulas the best that suits the question given in this case you can use v=u+at.since the time, acceleration and initial velocity have been given in the question

v=116+(9.2)13^2

v=116+119.6

=235.6m/s

I hope this helps

Why does the weight of a body differ on different celestial bodies of the universe?​

Answers

An object's weight is dependent on its mass and how strongly gravity pulls on it. The strength of gravity depends on how far away one object is from another.

what you filling your heart with
oxygen and blood

Answers

Answer:

Explanation:

                    The right side of your heart receives oxygen-poor blood from your veins and pumps it to your lungs, where it picks up oxygen and gets rid of carbon dioxide. The left side of your heart receives oxygen-rich blood from your lungs and pumps it through your arteries to the rest of your body.

        #I AM ILLITERATE

Ben works as a medical assistant. He needs to take a patient's vitals, but the patient is refusing to cooperate. He hasn't experienced this before, so he decides to ask a nurse for advice on how to handle it. This is making a decision by O a) delegation. O b) command. c) vote. O d) consult. Question

Solve the following numerical problems. a) A load of 400N is lifted up by an effort of 100N. If load distance is 20cm, what will be the effort distance? (Ans: 80cm) b) Two boys, Shrijan having weight 600N and Shrijesh having weight 300N are playing see-saw. If Shrijan is sitting at 2m from fulcrum, where should Shrijesh sit from fulcrum to balance Shrijan?(Ans: 4m) c) A lever of length 1m has been used ttoko lift a load of 600N by applying an effort of 200N. If load is at 20cm from fulcrum, calculate mechanical advantage, velocity ratio and efficiency. (Ans: MA = 3, VR = 4, n=75%) d) Study the figure below and find the value of effort. (Ans: 120N) Muyn) 2.5m 600N 0.5m ? Science and Enyin​

Answers

Answer:

given,

load = 400 N

effort = 100 N

load distance = 20 cm

we know that ,

E*Ed = L*Ld

=100 N* Ed = 400N * 20 cm

=100N * Ed = 8000N/cm

= Ed =( 8000N/cm ) / 100N

= Ed = 80 cm

b. soln.

given,

load = 600 N

load distance = 2 m

effort = 300 N

effort distance = ?

we know that ,

= E *Ed = L * Ld

= 300N * Ed = 600N * 2 m

= 300N * Ed = 1200N/m

=Ed =( 1200N/m ) / 300 N

= Ed = 4 m

C. soln.

given,

load = 600 N

load distance =20 cm

effort = 200 N

effort distance = ?

M.A = ?

V.R = ?

Efficiency = ?

we know that ,

= E *Ed = L *Ld

= 200N * Ed = 600 N * 20 cm

=200 N *Ed = 12000 N/cm

=Ed = ( 12000 N/cm) / 200 N

= Ed = 60 cm

Also,

M.A = load / effort

=600 N / 200 N

= 3

V.R = Ed/ Ld

= 60 cm / 20 cm

= 4

efficiency = ( M.A / V.R ) 100 %

= ( 3 / 4 ) 100%

= 75 %

d. soln.

given,

load = 600 N

load distance = 0.5 m

effort distance = 2.5 m

effort = ?

we know that ,

= E * Ed = L * Ld

= E * 2.5 m = 600 N * 0.5 m

= E * 2.5 m = 300 N / m

= E = ( 300 N / m ) / 2.5 m

= E = 120 N

i. The lift raises a car to a height of 1.8 m using a force of 5500 N. How much work does the lift
perform? (1 point)

Answers

Work = force x distance

Work = 5500 x 1.8

Work = 9900 N

The work measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement.

Work = force x distance

Work = 5500 x 1.8

Work = 9900 N

therefore, work does 9900 N

What is work?

Work is force applied over distance. Examples of work include lifting an object against the Earth's gravitation, driving a car up a hill, and pulling down a captive helium balloon. Work is a mechanical manifestation of energy. The standard unit of work is the joule (J), equivalent to a newton - meter (N · m).

What is work and energy?

Work is defined as transferring energy into an object so that there is some displacement. Energy is defined as the ability to do work. Work done is always the same. Energy can be of different types such as kinetic and potential energy.

To learn more about work, refer

https://brainly.com/question/19987285

#SPJ2

Why is velocity proportial to pressure?​

Answers

Answer:

Pressure and velocity are inversely proportional to each other because if pressure increase, the velocity decrease to keep the algebraic sum of potential energy, kinetic energy and pressure constant.

Which of the following processes occurs in a battery?

Batteries convert mechanical energy into chemical energy.
Batteries convert chemical energy into kinetic energy.
Batteries convert mechanical energy into electric energy.
Batteries convert chemical energy into electric energy.

Answers

Answer:

Batteries convert chemical energy into electric energy.

Explanation:

As seen, batteries have chemical energy in them stored which when used in any electronics produces electricity.

Thus, we can conclude that a battery converts chemical energy into electric energy.

What do is mean by environment friendly behaviour?​

Answers

Answer:
Being environmentally friendly means having a lifestyle that is better for the environment

Answer me as much as you can

Answers

Answer:

b) Fluorescent powder.

Calculate the efficiency of the following appliances:
1. A radiator that converts 1000) of electrical energy into 900J
of heat energy and 100J of light energy
2. A torch that converts 100J of chemical energy into 35) of
light energy and 65J of heat energy
3. A car that converts 10,000J of chemical energy into 6000) of
kinetic energy and 4000J of heat energy.
4. An energy saver light converts 1,000J of electrical energy
into 7003 of light energy and 300J of heat energy.
5. A speaker converts 100J of electrical energy into 50J of
sound energy and 50J of heat and kinetic energy.
we

Answers

Answer:

1. The efficiency of the radiator is 90 %

2. The efficiency of the torch is 65 %

3. The efficiency of the car is 40 %

4. The efficiency of the energy saver is 70 %

5. The efficiency of the speaker is 50 %

Explanation:

Efficiency = (Useful energy out ÷ Total energy in) × 100 J

1. Useful energy = 900 J

The total energy in = 1000 J

The efficiency of the radiator = ((900 J)/(1,000 J)) × 100 % = 90 %

2. Useful energy = 65 J

The total energy in = 100 J

The efficiency of the torch = ((65 J)/(100 J)) × 100 % = 65 %

3. Useful energy = 4,000 J

The total energy in = 10,000 J

The efficiency of the car = ((4,000 J)/(10,000 J)) × 100 % = 40 %

4. Useful energy = 700 J

The total energy in = 1,000 J

The efficiency of the energy saver = ((700 J)/(1,000 J)) × 100 % = 70 %

5. Useful energy = 50 J

The total energy in = 100 J

The efficiency of the speaker = ((50 J)/(100 J)) × 100 % = 50 %

mention various medium level and higher level human resources related to engineering.

Answers

Answer:

Higher-level human resources related to engineering are those who:

Assist other engineers with solving complex problems. Work with management to keep the project on track. Mentor younger engineers

Medium Level engineers will normally comprise the following:

They are responsible for developing engineering plans and performing calculations with respect to the same.  Project workflow management Engineering Support facilitation Project Reporting Coordination of Team Members

Cheers

Which court would you go to if you wanted to declare bankruptcy?
A.
Civil court
B.
U.S. Court of Appeals
C.
Supreme Court
D.
U.S. District Court

Answers

Answer:

A. Cilvil court.........

Civil court  would you go to if you wanted to declare bankruptcy. The correct option is A.

Thus, According to the United States Bankruptcy Code, bankruptcy proceedings are typically filed in a federal bankruptcy court.

The federal court system includes these courts, which deal with Chapter 7, Chapter 11, and Chapter 13 bankruptcy cases.

The additional alternatives (U.S. Court of Appeals, Supreme Court, and U.S. District Court) are not the main courts where bankruptcy cases are generally filed.

Thus, Civil court  would you go to if you wanted to declare bankruptcy. The correct option is A.

Learn more about Civil court, refer to the link:

https://brainly.com/question/29614273

#SPJ3

Discuss the role of globalization in the development of sI unit​

Answers

Answer:

Sharing of information

Explanation:

The development of SI unit has helped in the sharing of scientific as well as techical information internationally.

Answer:

It was created during the French Revolution in 1799 and has enabled for the international exchange of scientific and technical information. Calculating with SI units is also a lot easier than using the English system.

A train travels 600 kilometers in 1 hour. What is the train's velocity in meters/second?

Answers

here's the answer to your question

an athlete had lifts a load with a mass of 150kg.

1) calculate the gravitational potential energy gained

2) if the mass of the load is increased to 200kg, calculate the gravitational potential energy gained by the load.

3) based on answers in 1&2, state the relationship between the mass of the load and the gravitational potential energy.

Answers

Mass=150kg=mHeight=h=2mAcceleration due to gravity=g=10m/s^2

[tex]\\ \sf\longmapsto P.E=mgh[/tex]

[tex]\\ \sf\longmapsto P.E=150(2)(10)[/tex]

[tex]\\ \sf\longmapsto P.E=3000J[/tex]

In 2nd case

Mass =m=200kgHeight=h=2mg=10m/s^2

[tex]\\ \sf\longmapsto P.E=200(2)(10)[/tex]

[tex]\\ \sf\longmapsto P.E=4000J[/tex]

We can observe that

If mass of body increases gravitational potential energy will increase.

Pls help me with this fast. I will mark brainiest

Answers

Answer:

a) 70, 95

b) 95-70= 25cc

c) density= mass/volume

102/25

=4.08g/cc

define one standard kg ​

Answers

Answer:

The mass of platinum iridium rod whose diameter and height are equal an kept at international bureau of weight and measurement in Paris of France is called as one standard kg.

What is an effect of continental drift?

Answers

Answer:  An effect of continental drift is causing tectonic plates resting upon the convecting mantle to move which results in natural disasters like earthquakes, volcanic eruptions, and more.

A certain superconducting magnet in the form of a solenoid of length 0.28 m can generate a magnetic field of 7.0 T in its core when its coils carry a current of 80 A. The windings, made of a niobium-titanium alloy, must be cooled to 4.2 K. Find the number of turns in the solenoid.

Answers

Answer:

The number of turns in the solenoid is 19506.

Explanation:

Length of solenoid, L = 0.28 m

Magnetic field, B = 7 T

Current, I = 80 A

Let the number of turns is N.

The formula used to find the magnetic field is

[tex]B = \frac{\mu 0N I}{L}\\\\7 = \frac{4 \times 3.14\times 10^{-7}\times N\times 80}{0.28}\\\\N = 19506[/tex]

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