A magnetic field is passing through a loop of wire whose area is 0.015 m2. The direction of the magnetic field is parallel to the normal to the loop, and the magnitude of the field is increasing at the rate of 0.20 T/s. (a) Determine the magnitude of the emf induced in the loop. (b) Suppose the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in m^2/s) should the area be changed at the instant when B

Answers

Answer 1

This question is incomplete, the complete question is;

A magnetic field is passing through a loop of wire whose area is 0.015 m2. The direction of the magnetic field is parallel to the normal to the loop, and the magnitude of the field is increasing at the rate of 0.20 T/s.

(a) Determine the magnitude of the emf induced in the loop.

(b) Suppose the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in m^2/s) should the area be changed at the instant when B = 1.5 T, if the induced emf is to be zero? (Give the magnitude of the rate of change of the area.) (m2/s)

Answer:

a) the magnitude of the emf induced in the loop is 0.003 V

b) dA/dt = 0.002 m²/s

Explanation:

Area of the loop wire A = 0.015 m²

magnitude of the field is increasing dB/dt = 0.20 T/s

a)

Determine the magnitude of the emf induced in the loop.

V = A( dB/dt )

we substitute

V = 0.015 m² × 0.20 T/s

V = 0.003 V

Therefore,  the magnitude of the emf induced in the loop is 0.003 V

b)  the induced emf is;

V = B( dA/dt ) + A( dB/dt )

given that; induced emf is 0, B = 1.5

so we substitute

0 = [ 1.5T × ( dA/dt ) ] + [ 0.015 m² × 0.20 T/s ]

-[ 1.5T × ( dA/dt )] = 0.003 m²T/s

dA/dt = -[ 0.003 m²T/s / 1.5T ]

dA/dt = -0.002 m²/s

the negative shows that the area is decreasing

hence, dA/dt = 0.002 m²/s


Related Questions

The current through each resistor in the two-resistor circuit is _________ the current through the resistor in the one-resistor circuit (the circuit in Part A). The voltage across each resistor in the two-resistor circuit is ___________ the voltage across the resistor in the one-resistor circuit.

Answers

Answer:

Serial circuit.   the current is constant.  

            the voltage across a given resistor is half the rating in a one-resistor circuit.

Parallel circuit    the voltage is constant

the current is half the value of the current with a single resistor

Explanation:

To answer exactly this exercise, you need the diagram or the indication of the type of circuit being used, since there are two possibilities, let's consider the results of each one.

Serial circuit.

The two resistance are one after the other.

In this case the current in the resistor sides is the same, that is, the current is constant in the circuit.

The voltage is proportional to the value of each resistor and if the two resistors are equal, the voltage across a given resistor is half the rating in a one-resistor circuit.

Parallel circuit

the two resistance is next to each other.

In this case the voltage is constant, that is, the voltage across the two resistors is the same as in the case of a single resistor.

The current is inversely proportional to the value of the resistance

          i₁ = V / R₁

          i₂ = V / R₂

     

for a single resistance

           I = V / R

these currents are related

          i = i₁ + i₂

if the two resistors have the same value the current is half the value of the current with a single resistor

Hai quả cầu kim loại nhỏ giống nhau, mang điện tích q1 = 2.10-8 C; q2 = 6.10-8 C, đặt cách

nhau một đoạn r trong không khí thì chúng đẩy nhau bằng một lực là 18.10-5 N. Cho hai quả cầu

tiếp xúc nhau rồi đưa về khoảng cách cũ thì lực tương tác giữa hai quả cầu là:

Answers

Answer:

fgggggffgggcffghhhjjkuuu of to ok with queen size of your yyyygtyttttttyyyhgggghhhhfrghjjkkk

Explanation:

jjjgggyuuuuiiii hhjiiihuyyuuugggyujjhhhhggghhhjjhhhhjhhui

on a horizontal axis whose unit is the meter, a linear load ranging from 0 to 1 ma a linear load distribution = 2 nC / m.
determine the modulus of the electric field created by the previous loaded bar at the point A of abscissa 2m (we have to find the relation between l, which is the distance between the elementary bar and the point A and x which sweeps the segment [0: 1]

Answers

Answer:

The correct answer is - 8.99N/C

Explanation:

[tex]dE=k\dfrac{dq}{x^2}\\ dq=\lambda{dx}\\ \lambda=2nC/m\\ dq=2dxnC\\ dE=k\dfrac{2dx}{x^2}\\ E=2k\int_1^2\dfrac{dx}{x^2}\\ E=2k(\frac{-1}{x})_1^2=k\times10^{-9}N/C\\ E=8.99\times10^9\times10^{-9}N/C\\ E=8.99N/C\\dE=k[/tex]

what is environment in world​

Answers

Answer:

World Environment Day has been celebrated every year on 5 June, engaging governments, businesses and citizens in an effort to address pressing environmental issues

Explanation:

Answer:

The natural environment or natural world encompasses all living and non-living things occurring naturally, meaning in this case not artificial. The term is most often applied to the Earth or some parts of Earth.

A mass weighing 4 lb stretches a spring 4in. Suppose the mass is given an additional in displacement downwards and then released. Assuming no friction and no external force, the natural frequency W (measured in radians per unit time) for the system is? (Recall that the acceleration due to gravity is 32ft/sec2).
a) None of the other alternatives is correct.
b) W = v2 3
c)w=212
d) w = 4/6
e) w=213

Answers

Answer:

4√6 rad/s

Explanation:

Since the spring is initially stretched a length of x = 4 in when the 4 lb mass is placed on it, since it is in equilibrium, the spring force, F = kx equals the weight of the mass W = mg.

So, W = F

mg = kx where m = mass = 4lb, g = acceleration due to gravity = 32 ft/s², k = spring constant and x = equilibrium displacement of spring = 4 in = 4 in × 1ft /12 in = 1/3 ft

making k the spring constant subject of the formula, we have

k = mg/x

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

k = mg/x  

k = 4 lb × 32 ft/s² ÷ 1/3 ft

k = 32 × 4 × 3

k = 384 lbft²/s²

Now, assuming there is no friction and no external force, we have an undamped system.

So, the natural frequency for an undamped system, ω = √(k/m) where k = spring constant = 384 lbft²/s² and m = mass = 4 lb

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

ω = √(k/m)

ω = √(384 lbft²/s² ÷ 4 lb)

ω = √96

ω = √(16 × 6)

ω = √16 × √6

ω = 4√6 rad/s

A Geiger counter registers a count rate of 8,000 counts per minute from a sample of a radioisotope. The count rate 24 minutes later is 1,000 counts per minute. What is the half-life of the radioisotope?

Answers

11.54 minutes

Explanation:

The decay rate equation is given by

[tex]N = N_0e^{-\frac{t}{\lambda}}[/tex]

where [tex]\lambda[/tex] is the half-life. We can rewrite this as

[tex]\dfrac{N}{N_0} = e^{-\frac{t}{\lambda}}[/tex]

Taking the natural logarithm of both sides, we get

[tex]\ln \left(\dfrac{N}{N_0}\right) = -\left(\dfrac{t}{\lambda}\right)[/tex]

Solving for [tex]\lambda[/tex],

[tex]\lambda = -\dfrac{t}{\ln \left(\frac{N}{N_0}\right)}[/tex]

[tex]\:\:\:\:= -\dfrac{(24\:\text{minutes})}{\ln \left(\frac{1000\:\text{counts/min}}{8000\:\text{counts/min}}\right)}[/tex]

[tex]\:\:\:\:=11.54\:\text{minutes}[/tex]

Determine the acceleration of a pendulum bob as it passes through an angle of 15 degrees to the right of the equilibrium point.

Answers

Answer:

Explanation:

Since energy is conserved:

2

mu  

2

 

=  

2

mv  

2

 

+mgh

⇒u  

2

=v  

2

+2gh

⇒(3)  

2

=v  

2

+2(9.8)(0.5−0.5cos60)

⇒v=2m/s

Acceleration of the simple pendulum is 2.62 m/s².

What is meant by a simple pendulum ?

When a point mass is suspended from a fixed support by a light, non-extensible string, the instrument is said to be a simple pendulum.

Here,

Let the mass of the bob be m. The simple pendulum is attached to the fixed support with a string having length l. The pendulum makes an angle of 15° with the vertical from the equilibrium point.

Let T be the tension acting on the string.

As, the bob passes through the angle,

The weight of the bob becomes equal to the vertical component of the tension.

mg = T cos15°

Also, the horizontal component of the tension,

T sin15° = ma

By solving these two equations, we get that,

Acceleration of the simple pendulum,

a = g tan15°

a = 9.8 x 0.267

a = 2.62 m/s²

Hence,

Acceleration of the simple pendulum is 2.62 m/s².

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A 20 N south magnetic force pushes a charged particle traveling with a velocity of 4 m/s west through a 5 T magnetic field pointing downwards . What is the charge of the particle ?

Answers

Answer:

Charge of the particle is 1 coulomb.

Explanation:

Force, F:

[tex]{ \bf{F=BeV}}[/tex]

F is magnetic force.

B is the magnetic flux density.

e is the charge of the particle.

V is the velocity

[tex]{ \sf{20 = (5 \times e \times 4)}} \\ { \sf{20e = 20}} \\ { \sf{e = 1 \: coulomb}}[/tex]


The temperature of a body falls from 30°C to 20°C in 5 minutes. The air
temperature is 13°C. Find the temperature after a further 5 minutes.

Answers

Answer:

15.88

is the correct answer

The masses of two heavenly bodies are 2×10‘16’ and 4×10 ‘22’ kg respectively and the distance between than is 30000km. find the gravitational force between them ? ans. 2.668× 10-9N​

Answers

[tex]F = 5.93×10^{13}\:\text{N}[/tex]

Explanation:

Given:

[tex]m_1= 2×10^{16}\:\text{kg}[/tex]

[tex]m_2= 4×10^{22}\:\text{kg}[/tex]

[tex]r = 30000\:\text{km} = 3×10^7\:\text{m}[/tex]

Using Newton's universal law of gravitation, we can write

[tex]F = G\dfrac{m_1m_2}{r^2}[/tex]

[tex]\:\:\:\:=(6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2)\dfrac{(2×10^{16}\:\text{kg})(4×10^{22}\:\text{kg})}{(3×10^7\:\text{m})^2}[/tex]

[tex]\:\:\:\:= 5.93×10^{13}\:\text{N}[/tex]

a. Do the waves made by the two faucets travel faster than the waves made by just one faucet?
b. How do you know this? Describe how the two-faucet wave pattern compares with the one-faucet pattern.
c. Describe what happens to the two-faucet wave pattern as the separation of the faucets is increased.

Answers

Answer:

asdasd dsa dasdasd sadas dasd asdasd asd asd dsa asdd 223 aasd ada dasd sa dasd dsaa sd adsd asasd

Explanation:

point charges q1=50 uc and q2=-25 uc are placed 1 m apart. what is the force on a third chare q3=2 uc placed midway between q1 and q2? where must q3 of the preceding problem be placed so that the net force on it is zero?

Answers

Answer:

d = -1 m

The negative sign indicates that the charge from that force of the space of the two spheres.

Explanation:

That is a problem of electric forces, given by Coulomb's law

          F = [tex]k \frac{ q1q2}{r^2}[/tex]

We use that charges of the same sign repel and charges of different signs do not attract, so the net force is

           ∑ = F₁₃ + F₂₃

          F_ {net} = [tex]k \frac{q_1q_3}{r_{13}^2} + k \frac{q_2q_3}{ r_{23}^}[/tex]

a) the charge is placed at the midpoint between the other two

          r₁₃ = r₁₂ = R = ½ m = 0.5m

         F_ {net} =[tex]\frac{k}{R^2 } \ q3 ( q1+q2)[/tex]

calculate us

          F_ {net} = 9 10⁹ / 0.5²   2 10⁻⁶ (50 -25) 10⁻⁶

          F_ {net} = 1,800 N

b) where must be placed q3 so that the force is zero

for this case the charge q3 is outside the spheres

          ∑ F = 0

          F₁₁₃ = F₂₃

          k q_1 / r_{13}² = k q₂ q₃ / r₂₃²

          q₁/ r₁₂²   = q₂ / r₂₃²

suppose the distance

          r₁₂ = d

the he other sphere is

          r₂₃ = d + 1

             

we substitute

          q₃ / d² = q₂ / (d + 1) ²

          (d + 1) ² = q₂ / q₃ d²

           d² (1 - q₂/ q₃) + 2d + 1 = 0

we solve the equation of a second

            d = [-2 + [tex]\sqrt{2^2 - 4 1 ( 1+25/50}[/tex] ] / 2

             d = -2 /2

             d = -1 m

The negative sign indicates that the charge from that force of the space of the two spheres.

A 2.0 kg puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. How far does the puck move from rest in 2.25 s?

Answers

Answer:

d = 6.32 m

Explanation:

Given that,

The mass of a puck, m = 2 kg

It is pushed straight north with a constant force of 5N for 1.50 s and then let go.

We need to find the distance covered by the puck when move from rest in 2.25 s.

We know that,

F = ma

[tex]a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2[/tex]

Let d is the distance moved in 2.25 s. Using second equation of motion,

[tex]d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m[/tex]

So, it will move 6.32 m from rest in 2.25 seconds.

Use the DC Construction kit to build a simple circuit to perform the following task:

You are asked to use a single resistor and a 110 V DC battery for the purpose of boiling a litter of water (4,184 Joule/Kg*degree Celsius), with a starting temperature of 20 C, in exactly 4 minutes.

Answers

Answer:

The resistance is 8.7 ohm.

Explanation:

Voltage, V = 110 V

mass, m = 1 kg

change in temperature, T = 100 - 20 = 80 C

time, t = 4 min = 4 x 60 = 240 s

specific heat, c = 4184 J/kg C

let the resistance is R.

The heat generated by the heater is used to the heat the water.

[tex]\frac{V^2}{R} t = m c T \\\\\frac{110^2}{R}\times 240 = 1\times 4184\times 80\\\\R = 8.7 ohm[/tex]

How much amount of water can be decomposed
through electrolysis by passing 2 F charge?

Answers

Answer:

So, with 2 Faraday of electricity, we can decompose (2/4 × 2) = 1 mole of water. So 18 grams of water is decomposed.

a volcano that may erupt again at some time in the distant future is

Answers

The answer is a dormant volcano

An 1800-W toaster, a 1400-W electric frying pan, and a 55-W lamp are plugged into the same outlet in a 15-A, 120-V circuit. (The three devices are in parallel when plugged into the same socket.)

a. Will this combination blow the 15-A fuse?
b. What current is drawn by each device?

Answers

Being in parallel each device will have an equal voltage drop of 120 V

A. Yes the combination will blow the fuse. See part B for the total current.

B. Toaster = 1800W / 120V = 15A

   Frying Pan = 1400W / 120V = 11.67A

  Lamp = 55W / 120V = 0.458A

Total amps = 15 + 11.67 + 0.458 = 27.128 Amps

27.128A is greater than 15A so the fuse will blow.

A spherically mirrored ball is slowly lowered at New Years Eve as midnight approaches. The ball has a diameter of 8.0 ft. Assume you are standing directly beneath it and looking up at the ball. When your reflection is half your size then the mirror is _______ ft above you.

Answers

Answer:

The distance between mirror and you is 2 ft.

Explanation:

diameter, d = 8 ft

radius of curvature, R = 4 ft

magnification, m = 0.5

focal length, f = R/2 = 4/2 = 2 ft

let the distance of object is u and the distance of image is v.

[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\\\frac{1}{2}=\frac{1}{v}+\frac{1}{u}\\\\v = \frac {2 u}{u - 2}[/tex]

Use the formula of magnification

[tex]m = \frac{v}{u}\\\\0.5 =\frac { u}{u - 2}\\ \\u - 2 = 2 u \\\\u = -2 ft[/tex]

What do scientists use to determine the temperature of a star?

Answers

Answer:

Measure the brightness of a star through two filters and compare the ratio of red to blue light. Compare to the spectra of computer models of stellar spectra of different temperature and develop an accurate color-temperature relation.

ai là người phát hiện trái đất hình cầu đầu tiên ?

Answers

Answer:

Can't understand the language

how interfacial angles are determined using contact goinometer​

Answers

Interfacial angles can be measured using a contact goniometer or more precisely using a reflection goniometer. See Klein & Hurlbut's figure 2.40. The interfacial angle between two faces of a mineral crystal is identical for all crystals of the same mineral that exhibit the corresponding two faces.

A 2kg ball is rolled along the floor for 0.8 m at a constant speed of 6 m/s. What is the work done by gravity?

A, 0
B, 16 J
C, 72 J
D, 450 J
E, 90 J

Answers

=F×s×cosa=2×g×0,8×cos90°= 0

The work done by gravity on a ball of 2 kg which is moving with a constant speed of 6 meter per second is zero. Thus, the correct option is A.

What is Work?

Work is the energy transfer to or from an object through the application of force along with the displacement. For a constant force aligned with the direction of motion, the work done is equal to the product of the force strength which is applied and the distance traveled by the object.

Work = Force × Displacement

Force = Mass × Acceleration

Acceleration of the ball is zero as it is moving with a constant speed. Therefore, the work done by the gravity is zero.

Therefore, the correct option is A.

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Question
Name and write briefly on the international body
that Introduced si units .

Answers

Answer:

International System of Units. it was established in 1960 by the 11the General Conference on Weights and Measures

In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue ball bounces off with a speed of 0.8 m/s at an angle of 20', as shown in the diagram below. Both balls have a mass of 0.6 kg.
a) what is the momentum of the system before the collision
b) what is the momentum after the collision
c) what angle dose the right ball travel after the collision
d) what is the magnitude of the eight balls velocity after the collision

Answers

Answer:

a)  p₀ = 1.2 kg m / s,  b) p_f = 1.2 kg m / s,  c)   θ = 12.36, d)  v_{2f} = 1.278 m/s

Explanation:

For this exercise we define a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved

 

a) the initial impulse is

         p₀ = m v₁₀ + 0

         p₀ = 0.6 2

         p₀ = 1.2 kg m / s

b) as the system is isolated, the moment is conserved so

        p_f = 1.2 kg m / s

we define a reference system where the x-axis coincides with the initial movement of the cue ball

 

we write the final moment for each axis

X axis

         p₀ₓ = 1.2 kg m / s

         p_{fx} = m v1f cos 20 + m v2f cos θ

         p₀ = p_f

        1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ

         1.2482 = v_{2f} cos θ

Y axis  

        p_{oy} = 0

        p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ

        0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ

        0.2736 = v_{2f} sin θ

we write our system of equations

         0.2736 = v_{2f} sin θ

         1.2482 = v_{2f} cos θ

divide to solve

         0.219 = tan θ

          θ = tan⁻¹ 0.21919

          θ = 12.36

let's look for speed

            0.2736 = v_{2f} sin θ

             v_{2f} = 0.2736 / sin 12.36

            v_{2f} = 1.278 m / s

Steel railway tracks are laid at 8oC. What size of expansion gap are needed 10m long rail sections if the ambient temperature varies from -10oC to 50oC? [Linear expansivity of steel = 12 x]​

Answers

Answer:

Gap left = Change in length on heating

Gap=Initial length×Coefficient of linear expansion×change in temperature

Gap=10×0.000012×15m

⟹Gap=0.0018 m

this is an example u have to put your equation in it

Assume that a friend hands you a 10-newton box to hold for her. If you hold the box without moving it at a height of 10 meters above the ground, how much work do you do

Answers

Answer:

100 Joules

Explanation:

Applying,

W = mgh................... Equation 1

Where W = workdone to hold the box above the ground, mg = weight of the box, h = height of the box.

From the question,

Given: mg = 10 newtons, h = 10 meters.

Substitute these values into equation 1

W = 10×10

W = 100 Joules.

Hence the amount of workdone is 100 Joules

Một loa phát ra với cường độ âm là 40 (W/m2

). Mức cường độ âm của loa thuộc phạm vi?

Answers

Answer:ew

Explanation:

qeeqw

A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1° from the normal to the plane of the loop. The magnitude of the magnetic field is 3.35 T , and the radius of the wire loop is 0.240 m . Find the magnetic flux Φ through the loop.

Answers

Answer:

0.5849Weber

Explanation:

The formula for calculating the magnetic flus is expressed as:

[tex]\phi = BAcos \theta[/tex]

Given

The magnitude of the magnetic field B = 3.35T

Area of the loop = πr² = 3.14(0.24)² = 0.180864m²

angle of the wire loop θ = 15.1°

Substitute the given values into the formula:

[tex]\phi = 3.35(0.180864)cos15.1^0\\\phi =0.6058944cos15.1^0\\\phi =0.6058944(0.9655)\\\phi = 0.5849Wb[/tex]

Hence the magnetic flux Φ through the loop is 0.5849Weber

5. A bicyclist is finishing her repair of a flat tire when a friend rides by at a constant velocity of
3.5 m/s. Three seconds later, the bicyclist hops on her bike and accelerates at 3.6 m/s² until she
catches her friend.
a. How much time does it take until she catches her friend?
b. How far has she traveled in this time?
c. What is her speed when she catches up?

Answers

Answer:

a) t = 3.6 s

b) d = 23 m

c) v = 13 m/s

Explanation:

Let t be the time the accelerating rider rides

the distance she travels is

d = ½3.6t²

the distance for the other cyclist is

d =3.5(t + 3)

½3.6t² = 3.5(t + 3)

1.8t² - 3.5t - 10.5 = 0

quadratic formula, positive answer

t = (3.5 + √(3.5² - 4(1.8)(-10.5))) / (2(1.8))

t = 3.575786...

d = ½(3.6)(3.575786²) = 23.015...

v = 3.6(3.575786) = 12.8728...

Assume the speed of sound is 343 m/s. You are sitting 150 m away from home plate at a baseball game. How much time in seconds elapses between the batter hitting a home run and the moment you actually hear the batter hitting the ball

Answers

Answer:

  t = 0.437 s

Explanation:

Sound is a wave so its speed is constant

         v = x / t

         t = x / v

         

indicates that the distance is x = 150 m

         t = 150/343

         t = 0.437 s

this is the time it takes to hear the hit

To see the blow it is almost instantaneous since the speed of light is much greater c = 3 10⁸ m / s

Other Questions
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