A fan spins at 6.0 rev/s. You turn it off, and it slows at 1.0 rev/s2. What is the angular displacement before it stops

Answers

Answer 1

Answer:

Angular displacement before it stops = 18 rev

Explanation:

Given:

Speed of fan w(i) = 6 rev/s

Speed of fan (Slow) ∝ = 1 rev/s

Final speed of fan w(f) = 0 rev/s

Find:

Angular displacement before it stops

Computation:

w(f)² = w(i) + 2∝θ

0² = 6² + 2(1)θ

0 = 36 + 2θ

2θ = -36

Angular displacement before it stops = -36 / 2

θ = -18

Angular displacement before it stops = 18 rev


Related Questions

Imagine a spaceship traveling at a constant speed through outer space. The length of the ship, as measured by a passenger aboard the ship, is 28.2 m. An observer on Earth, however, sees the ship as contracted by 16.1 cm along the direction of motion. What is the speed of the spaceship with respect to the Earth

Answers

[tex]3.20×10^7\:\text{m/s}[/tex]

Explanation:

Let

[tex]L = 28.2\:\text{m}[/tex]

[tex]L' = 28.2\:\text{m} - 0.161\:\text{m} = 28.039\:\text{m}[/tex]

The Lorentz length contraction formula is given by

[tex]L' = L\sqrt {1 - \left(\dfrac{v^2}{c^2}\right)}[/tex]

where L is the length measured by the moving observer and L' is the length measured by the stationary Earth-based observer. We can rewrite the above equation as

[tex]\sqrt {1 - \left(\dfrac{v^2}{c^2}\right)} = \dfrac{L'}{L}[/tex]

Taking the square of the equation, we get

[tex]1 - \left(\dfrac{v^2}{c^2}\right) = \left(\dfrac{L'}{L}\right)^2[/tex]

or

[tex]1 - \left(\dfrac{L'}{L}\right)^2 = \left(\dfrac{v}{c}\right)^2[/tex]

Solving for v, we get

[tex]v = c\sqrt{1 - \left(\dfrac{L'}{L}\right)^2}[/tex]

[tex]\:\:\:\:=(3×10^8\:\text{m/s})\sqrt{1 - \left(\dfrac{28.039\:\text{m}}{28.2\:\text{m}}\right)^2}[/tex]

[tex]\:\:\:\:=3.20×10^7\:\text{m/s} = 0.107c[/tex]

Serena wants to play a trick on her friend Marion. She adds salt, sugar, and vinegar into her glass of water when Marion is out of the room. Why does she know that Marion will drink the water?

Answers

Maybe because they’re friends and the stuff she put in there will become clear again, leaving the water clear and the friend without any suspicions?

A force cannot exist without an agent and a system.
True
False

Answers

Answer:

true

Explanation:

forces require an agent

you should always be able to identify what (the agent) is producing the force

M
A boy of mass 60 kg and a girl of mass 40 kg are
together and at rest on a frozen pond and push
each other apart. The girl moves in a negative
direction with a speed of 3 m/s. What must be the
total final momentum of the boy AND girl
combined?
A. -120 kgm/s
B. 0 kgm/s
C. -100 kgm/s
D. 120 kgm/s

Answers

Answer:

option D thinking so

Explanation:

okay na your whish

A light beam inside a container of some liquid hits the surface of the liquid/air interface. Depending on the angle, the light beam may or may not be able to get out into the air medium. At what angle will the beam of light be totally reflected back into the liquid in the container

Answers

Answer:

Following are the solution to the given question:

Explanation:

Applying the Snell Law:

when is the liquid's refractive index, is the air's refractive index.  

the liquid's refractive index,   the air's index of refraction.

It is the limiting case when , Inside the interphase of two mediums, light is scattered. Thus,

[tex]n_l \sin \theta_l = n_a \sin 90^\circ = n_a[/tex]

[tex]\theta_l = \arcsin \dfrac{n_a}{n_l} =\arcsin \dfrac{1}{1.38} = 46.4^\circ[/tex]

From  the incident angles [tex]\theta_l[/tex] is greater than 46.4°, that is the light reflected back into the liquid.

The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3. Express your answer numerically in joules.

Answers

The question is incomplete. The complete question is :

A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2 .

Find the energy U1 of the dielectric-filled capacitor. I got U1=2.99*10^-10 J which I know is correct. Now I need these:

1. The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.

2. The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3.

Solution :

Given :

[tex]A = 10 \ cm^2[/tex]

   [tex]$=0.0010 \ m^2$[/tex]

d = 10 mm

  = 0.010 m

Then, Capacitance,

[tex]$C=\frac{k \epsilon_0 A}{d}$[/tex]

[tex]$C=\frac{8.85 \times 10^{12} \times 3 \times 0.0010}{0.010}$[/tex]

[tex]$C=2.655 \times 10^{12} \ F$[/tex]

[tex]$U_1 = \frac{1}{2}CV^2$[/tex]

[tex]$U_1 = \frac{1}{2} \times 2.655 \times 10^{-12} \times (15V)^2$[/tex]

[tex]$U_1=2.987 \times 10^{-10}\ J$[/tex]

Now,

[tex]$C_k=\frac{1}{2} \frac{k \epsilon_0}{d} \times \frac{A}{2}$[/tex]

And

[tex]$C_{air}=\frac{1}{2} \frac{\epsilon_0}{d} \times \frac{A}{2}$[/tex]

In parallel combination,

[tex]$C_{eq}= C_k + C_{air}$[/tex]

[tex]$C_{eq} = \frac{1}{2} \frac{\epsilon_0 A}{d}(1+k)$[/tex]

[tex]$C_{eq} = \frac{1}{2} \times \frac{8.85 \times 10^{-12} \times 0.0010}{0.01} \times (1+3)$[/tex]

[tex]$C_{eq} = 1.77 \times 10^{-12}\ F$[/tex]

Then energy,

[tex]$U_2 =\frac{1}{2} C_{eq} V^2$[/tex]

[tex]$U_2=\frac{1}{2} \times 1.77 \times 10^{-12} \times (15V)^2$[/tex]

[tex]$U_2=1.99 \times 10^{-10} \ J$[/tex]

b). Now the charge on the [tex]\text{capacitor}[/tex] is :

[tex]$Q=C_{eq} V$[/tex]

[tex]$Q = 1.77 \times 10^{-12} \times 15 V$[/tex]

[tex]$Q = 26.55 \times 10^{-12} \ C$[/tex]

Now when the capacitor gets disconnected from battery and the [tex]\text{dielectric}[/tex] is slowly [tex]\text{removed the rest}[/tex] of the way out of the [tex]\text{capacitor}[/tex] is :

[tex]$C_3=\frac{A \epsilon_0}{d}$[/tex]

[tex]$C_3 = \frac{0.0010 \times 8.85 \times 10^{-12}}{0.01}$[/tex]

[tex]$C_3=0.885 \times 10^{-12} \ F$[/tex]

[tex]$C_3 = 0.885 \times 10^{-12} \ F$[/tex]

Without the dielectric,

[tex]$U_3=\frac{1}{2} \frac{Q^2}{C}$[/tex]

[tex]$U_3=\frac{1}{2} \times \frac{(25.55 \times 10^{-12})^2}{0.885 \times 10^{-12}}$[/tex]

[tex]$U_3=3.98 \times 10^{-10} \ J$[/tex]

Two resistances, R1 and R2, are connected in series across a 12-V battery. The current increases by 0.500 A when R2 is removed, leaving R1 connected across the battery. However, the current increases by just 0.250 A when R1 is removed, leaving R2 connected across the battery.
(a) Find R1.
Ω
(b) Find R2.
Ω

Answers

Answer:

a)   R₁ = 14.1 Ω,   b)  R₂ =  19.9 Ω

Explanation:

For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances

all resistors connected

           V = i (R₁ + R₂)

with R₁ connected

           V = (i + 0.5) R₁

with R₂ connected

           V = (i + 0.25) R₂

We have a system of three equations with three unknowns for which we can solve it

We substitute the last two equations in the first

           V = i ( [tex]\frac{V}{ i+0.5} + \frac{V}{i+0.25}[/tex] )

           1 = i ( [tex]\frac{1}{i+0.5} + \frac{1}{i+0.25}[/tex] )

           1 = i ( [tex]\frac{i+0.5+i+0.25}{(i+0.5) \ ( i+0.25) }[/tex] ) =  [tex]\frac{i^2 + 0.75i}{i^2 + 0.75 i + 0.125}[/tex]

           i² + 0.75 i + 0.125 = 2i² + 0.75 i

           i² - 0.125 = 0

           i = √0.125

           i = 0.35355 A

with the second equation we look for R1

          R₁ = [tex]\frac{V}{i+0.5}[/tex]

          R₁ = 12 /( 0.35355 +0.5)

          R₁ = 14.1 Ω

with the third equation we look for R2

          R₂ = [tex]\frac{V}{i+0.25}[/tex]

          R₂ =[tex]\frac{12}{0.35355+0.25}[/tex]

          R₂ =  19.9 Ω

An astronaut is traveling in a space vehicle that has a speed of 0.480c relative to Earth. The astronaut measures his pulse rate at 78.5 per minute. Signals generated by the astronaut's pulse are radioed to Earth when the vehicle is moving perpendicularly to a line that connects the vehicle with an Earth observer. (Due to vehicle's path there will be no Doppler shift in the signal.)
(a) What pulse rate does the Earth-based observer measure? beats/min
(b) What would be the pulse rate if the speed of the space vehicle were increased to 0.940c?
beats/min

Answers

Explanation:

The heart rate of the astronaut is 78.5 beats per minute, which means that the time between heart beats is 0.0127 min. This will be the time t measured by the moving observer. The time t' measured by the stationary Earth-based observer is given by

[tex]t' = \dfrac{t}{\sqrt{1 - \left(\dfrac{v^2}{c^2}\right)}}[/tex]

a) If the astronaut is moving at 0.480c, the time t' is

[tex]t' = \dfrac{0.0127\:\text{min}}{\sqrt{1 - \left(\dfrac{0.2304c^2}{c^2}\right)}}[/tex]

[tex]\:\:\:\:=0.0145\:\text{min}[/tex]

This means that time between his heart beats as measured by Earth-based observer is 0.0145 min, which is equivalent to 69.1 beats per minute.

b) At v = 0.940c, the time t' is

[tex]t' = \dfrac{0.0127\:\text{min}}{\sqrt{1 - \left(\dfrac{0.8836c^2}{c^2}\right)}}[/tex]

[tex]\:\:\:\:=0.0372\:\text{min}[/tex]

So at this speed, the astronaut's heart rate is 1/(0.0372 min) or 26.9 beats per minute.

Michelson and Morley concluded from the results of their experiment that Group of answer choices the experiment was successful in not detecting a shift in the interference pattern. the experiment was a failure since they detected a shift in the interference pattern. the experiment was a failure since there was no detectable shift in the interference pattern. the experiment was successful in detecting a shift in the interference pattern.

Answers

Answer:

The results of the experiment indicated a shift consistent with zero, and certainly less than a twentieth of the shift expected if the Earth's velocity in orbit around the sun was the same as its velocity through the ether.

Explanation:

Three children are riding on the edge of a merry-go-round that is a solid disk with a mass of 102 kg and a radius of 1.53 m. The merry-go-round is initially spinning at 9.71 revolutions/minute. The children have masses of 31.7 kg, 29.0 kg and 31.9 kg. If the child who has a mass of 29.0 kg moves to the center of the merry-go-round, what is the new angular velocity in revolutions/minute

Answers

Three children of masses and their position on the merry go round

M1 = 22kg

M2 = 28kg

M3 = 33kg

They are all initially riding at the edge of the merry go round

Then, R1 = R2 = R3 = R = 1.7m

Mass of Merry go round is

M =105kg

Radius of Merry go round.

R = 1.7m

Angular velocity of Merry go round

ωi = 22 rpm

If M2 = 28 is moves to center of the merry go round then R2 = 0, what is the new angular velocity ωf

Using conservation of angular momentum

Initial angular momentum when all the children are at the edge of the merry go round is equal to the final angular momentum when the second child moves to the center of the merry go round  Then,

L(initial) = L(final)

Ii•ωi = If•ωf

So we need to find the initial and final moment of inertia

NOTE: merry go round is treated as a solid disk then I= ½MR²

I(initial)=½MR²+M1•R²+M2•R²+M3•R²

I(initial) = ½MR² + R²(M1 + M2 + M3)

I(initial) = ½ × 105 × 1.7² + 1.7²(22 + 28 + 33)

I(initial) = 151.725 + 1.7²(83)

I(initial) = 391.595 kgm²

Final moment of inertial when R2 =0

I(final)=½MR²+M1•R²+M2•R2²+M3•R²

Since R2 = 0

I(final) = ½MR²+ M1•R² + M3•R²

I(final) = ½MR² + (M1 + M3)• R²

I(final)=½ × 105 × 1.7² + ( 22 +33)•1.7²

I(final) = 151.725 + 158.95

I(final) = 310.675 kgm²

Now, applying the conservation of angular momentum

L(initial) = L(final)

Ii•ωi = If•ωf

391.595 × 22 = 310.675 × ωf

Then,

ωf = 391.595 × 22 / 310.675

ωf = 27.73 rpm

Answer: So, the final angular momentum is 27.73 revolution per minute

1.An elevator is ascending with constant speed of 10 m/s. A boy in the elevator throws a ball upward at 20 m/ a from a height of 2 m above the elevator floor when the elevator floor when the elevator is 28 m above the ground.
a. What's the maximum height?
b. How long does it take for the ball to return to the elevator floor?​

Answers

(a) The maximum height reached by the ball from the ground level is 75.87m

(b) The time taken for the ball to return to the elevator floor is 2.21 s

The given parameters include:

constant velocity of the elevator, u₁ = 10 m/sinitial velocity of the ball, u₂ = 20 m/sheight of the boy above the elevator floor, h₁ = 2 mheight of the elevator above the ground, h₂ = 28 m

To calculate:

(a) the maximum height of the projectile

total initial velocity of the projectile = 10 m/s + 20 m/s  = 30 m/s (since the elevator is ascending at a constant speed)

at maximum height the final velocity of the projectile (ball), v = 0

Apply the following kinematic equation to determine the maximum height of the projectile.

[tex]v^2 = u^2 + 2(-g)h_3\\\\where;\\\\g \ is \ the \ acceleration \ due \ to\ gravity = 9.81 \ m/s^2\\\\h_3 \ is \ maximum \ height \ reached \ by \ the \ ball \ from \ the \ point \ of \ projection\\\\0 = u^2 -2gh_3\\\\2gh_3 = u^2 \\\\h_3 = \frac{u^2}{2g} \\\\h_3 = \frac{(30)^2}{2\times 9.81} \\\\h_3 = 45.87 \ m[/tex]

The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection

h = h₁ + h₂ + h₃

h = 28 m + 2 m  +  45.87 m

h = 75.87 m

(b) The time taken for the ball to return to the elevator floor

Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m

Apply the following kinematic equation to determine the time to return to the elevator floor.

[tex]h = vt + \frac{1}{2} gt^2\\\\where;\\\\v \ is \ the \ initial \ velocity \ of \ the \ ball \ at \ the \ maximum \ height = 0\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g}} \\\\t = \sqrt{\frac{2\times 47.87}{9.81}} \\\\t = 2.21 \ s[/tex]

To learn more about projectile calculations please visit: https://brainly.com/question/14083704

Mass A, 2.0 kg, is moving with an initial velocity of 15 m/s in the x-direction, and it collides with mass M, 4.0 kg, initially moving at 7.0 m/s in the x-direction. After the collision, the two objects stick together and move as one. What is the change in kinetic energy of the system as a result of the collision, in joules

Answers

Answer:

the change in the kinetic energy of the system is -42.47 J

Explanation:

Given;

mass A, Ma = 2 kg

initial velocity of mass A, Ua = 15 m/s

Mass M, Mm = 4 kg

initial velocity of mass M, Um = 7 m/s

Let the common velocity of the two masses after collision = V

Apply the principle of conservation of linear momentum, to determine the final velocity of the two masses;

[tex]M_aU_a + M_mU_m = V(M_a + M_m)\\\\(2\times 15 )+ (4\times 7) = V(2+4)\\\\58 = 6V\\\\V = \frac{58}{6} = 9.67 \ m/s[/tex]

The initial kinetic of the two masses;

[tex]K.E_i = \frac{1}{2} M_aU_a^2 \ + \ \frac{1}{2} M_mU_m^2\\\\K.E_i = (0.5 \times 2\times 15^2) \ + \ (0.5 \times 4\times 7^2)\\\\K.E_i = 323 \ J[/tex]

The final kinetic energy of the two masses;

[tex]K.E_f = \frac{1}{2} M_aV^2 \ + \ \frac{1}{2} M_mV^2\\\\K.E_f = \frac{1}{2} V^2(M_a + M_m)\\\\K.E_f = \frac{1}{2} \times 9.67^2(2+ 4)\\\\K.E_f = 280.53 \ J[/tex]

The change in kinetic energy is calculated as;

[tex]\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 280.53 \ J \ - \ 323 \ J\\\\\Delta K.E = -42.47 \ J[/tex]

Therefore, the change in the kinetic energy of the system is -42.47 J

A very long straight wire carries a 12 A current eastward and a second very long straight wire carries a 14 A current westward. The wires are parallel to each other and are 42 cm apart. Calculate the force on a 6.4 m length of one of the wires.

Answers

Answer: 5.12x10∧-4N

Explanation:

Force = I B L

L = 6.4m

Let Current (I) I₁ = I₂= 14A

Distance of the wire = 42cm = 0.42m

BUT

B = μ₀I / 2πr

=(2X10∧-7 X 12) / 0.42

       B =5.714×10∧-6T

Force = I B L

Force = 14x [5.714×10-6]×6.4

Force = 5.12x10∧-4N

Unit of speed is a derived unit. Give reasons​

Answers

Answer:

as it 8s based upon to fundamental units distance and Time

đổi đơn vị
42 ft2/hr to cm2/s

Answers

Answer:

X = 10.8387 cm²/s

Explanation:

In this exercise, you're required to convert a value from one unit to another.

Converting 42 ft²/hr to cm²/s;

Conversion:

1 ft² = 929.03 cm²

42 ft² = X cm²

Cross-multiplying, we have;

X = 42 * 929.03

X = 39019.26 cm²

Next, we would divide by time in seconds.

1 hour = 3600 seconds

X = 39019.26/3600

X = 10.8387 cm²/s

which energy does a car travelling 30 m/ph as it slows have:

a). chemical energy
b). thermal energy
c). kinetic energy

please helpp

Answers

Answer:

c) kinetic energy

Explanation:

Answer: C)  kinetic energy

Explanation:

Define simple harmonic motion. Write down the expressions for the velocity and aceraletion of such motion st different position along its path

Answers

Answer:

Simple harmonic motion is a special type of periodic motion or oscillation motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacemen

What is utilization of energy

Answers

Explanation:

Energy utilization focuses on technologies that can lead to new and potentially more efficient ways of using electricity in residential, commercial and industrial settings—as well as in the transportation sector

In the Bohr model of the hydrogen atom, an electron in the 3rd excited state moves at a speed of 2.43 105 m/s in a circular path of radius 4.76 10-10 m. What is the effective current associated with this orbiting electron

Answers

Answer:

Current =,charge / time

Charge = e = 1.6E-19 coulombs

t = T time for 1 revolution  (period)

v = S / T = distance traveled in 1 revolution / time for 1 revolution

T = S / v = 2 pi * 4.76E-10 / 2.43E5 = 1.23E-14

I = Q / T = 1.6E-19 / 1.23E-14 = 1.30E-5

What is the largest known star?

Answers

Answer:

UY Scuti is slightly larger than VY Canis Majoris

Explanation:

These stars are millions of miles away and cannot be seen by the naked eye.

Beetlejuice is another large star that can be seen by the eye.

Identify 2 different ways that data can be displayed or represented.​

Answers

Answer:

tables, charts and graphs

Explanation:

A generator uses a coil that has 270 turns and a 0.48-T magnetic field. The frequency of this generator is 60.0 Hz, and its emf has an rms value of 120 V. Assuming that each turn of the coil is a square (an approximation), determine the length of the wire from which the coil is made.

Answers

Answer:

The total length of wire is 0.24 m.

Explanation:

Number of turns, N = 270

magnetic field, B = 0.48 T

frequency, f = 60 Hz

rms value of emf = 120 V

maximum value of emf, Vo = 1.414 x 120 = 169.68 V

let the area of square is A and the side is L.

The maximum emf is given by

Vo = N B A w

169.68 = 270 x 0.48 x A x 2 x 3.14 x 60

A = 3.5 x 10^-3 m^2

So,

L = 0.0589 m

Total length of wire, P = 4 L = 4 x 0.0589 = 0.24 m

If at a particular instant and at a certain point in space the electric field is in the x-direction and has a magnitude of 3.10 V/mV/m , what is the magnitude of the magnetic field of the wave at this same point in space and instant in time

Answers

Answer:

B = 1.03 10⁻⁸ T

Explanation:

For an electromagnetic wave, the electric and magnetic fields must oscillate in phase so that they remain between them at all times, otherwise the wave will extinguish

       

This relational is expressed by the relation

           E /B = c

           B = E / c

let's calculate

            B = 3.10 / 3 10⁸

            B = 1.03 10⁻⁸ T

An energy efficient light bulb uses 15 W of power for an equivalent light output of a 60 W incandescent light bulb. How much energy is saved each month by using the energy efficient light bulb instead of the incandescent light bulb for 4 hours a day? Assume that there are 30 days in one month
A. 7.2 kW⋅hr
B. 21.6 kW⋅hr
C. 1.8 kW⋅hr
D. 5.4 kW⋅hr
E. 1.35 kW⋅hr

Answers

Answer: (d)

Explanation:

Given

15 W is equivalent to 60 W light that is, it save 45 W

So, for 4 hours it is, [tex]4\times 45=180\ W.hr[/tex]

For 30 days, it becomes

[tex]\Rightarrow 180\times 30=5400\ W.hr\\\Rightarrow 5.4\ kWh[/tex]

Thus, [tex]5.4\ kWh[/tex] is saved in 30 days

option (d) is correct.

the 2kg block slids down a firctionless curved ramp starting from rest at heiht of 3m what is the speed of the block at the bottemvof the ramp​

Answers

A

Explanation:

1qdeeeeeeeeeeehhhhhhhhhwilffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff.

Both of these questions are the same but their answers in the answer key are different. Why?

Answers

the person making the assignment must’ve made a mistake.

Explain why liquid particles at a high pressure would need more
energy to change to a gas than liquid particles at a low pressure.

Answers

Answer:

Liquids evaporate faster as they heat up and more particles have enough energy to break away. The particles need energy to overcome the attractions between them. ... At this point the liquid is boiling and turning to gas. The particles in the gas are the same as they were in the liquid they just have more energy.

What is cubical expansivity of liquid while freezing

Answers

Answer:

"the ratio of increase in the volume of a solid per degree rise of temperature to its initial volume" -web

Explanation:

tbh up above ✅

Answer:

cubic meter

Explanation:

Increase in volume of a body on heating is referred to as volumetric expansion or cubical expansion

A metal blade of length L = 300 cm spins at a constant rate of 17 rad/s about an axis that is perpendicular to the blade and through its center. A uniform magnetic field B = 4.0 mT is perpendicular to the plane of rotation. What is the magnitude of the potential difference (in V) between the center of the blade and either of its ends?

Answers

We are being given that:

The length of a metal blade = 300 cmThe angular velocity at which the metal blade is rotating about its axis is ω = 17 rad/sThe magnetic field (B) = 4.0 mT

A pictorial view showing the diagrammatic representation of the information given in the question is being attached in the image below.

From the attached image below, the potential difference across the conducting element of the length (dx) moving with the velocity (v) appears to be perpendicular to the magnetic field (B).

The magnitude of the potential difference induced between the center of the blade in relation to either of its ends can be determined by using the derived formula from Faraday's law of induction which can be expressed as:

[tex]\mathsf{E = B\times l\times v}[/tex]

where;

B = magnetic fieldl = lengthv = relative speed

From the diagram, Let consider the length of the conducting element (dx) at a distance of length (x) from the center O.

Then, the velocity (v) = ωx

The potential difference across it can now be expressed as:

[tex]\mathsf{dE = B*(dx)*(\omega x)}[/tex]

For us to determine the potential difference, we need to carry out the integral form from center point O to L/2.

[tex]\mathsf{E = \int ^{L/2}_{0}* B (\omega x ) *(dx)}[/tex]

[tex]\mathsf{E = B (\omega ) \times \Big[ \dfrac{x^2}{2}\Big]^{L/2}_{0}}[/tex]

[tex]\mathsf{E = B (\omega ) * \Big[ \dfrac{L^2}{8}\Big]}[/tex]

Recall that,

magnetic field (B) = 4 mT = 4 × 10⁻³  TLength L = 300 cm = 3mangular velocity (ω) = 17 rad/s

[tex]\mathsf{E = (4\times 10^{-3}) * (17) \Big[ \dfrac{(1.5)^2}{8}\Big]}[/tex]

[tex]\mathsf{E = 19.13 mV}[/tex]

Thus, we can now conclude that the magnitude of the potential difference as a result of the rotation caused by the metal blade from the center to either of its ends is 19.13 mV.

Learn more about Faraday's law of induction here:

https://brainly.com/question/13369951?referrer=searchResults

A very long, straight solenoid with a diameter of 3.00 cm is wound with 40 turns of wire per centimeter, and the windings carry a current of 0.235 A. A second coil having N turns and a larger diameter is slipped over the solenoid so that the two are coaxial. The current in the solenoid is ramped down to zero over a period of 0.40 s.

Required:
a. What average emf is induced in the second coil if it has a diameter of 3.5 cm and N = 7?
b. What is the induced emf if the diameter is 7.0 cm and N = 10?

Answers

Answer:

a) ε = 14.7 μv

b) ε = 21 μv

Explanation:

Given the data in the question;

Diameter of solenoid; d = 3 cm

radius will be half of diameter,  so, r = 3 cm / 2 = 1.5 cm = 1.5 × 10⁻² m

Number of turns; N = 40 turns per cm = 4000 per turns per meter

Current; [tex]I[/tex] = 0.235 A

change in time Δt = 0.40 sec

Now,

We determine the magnetic field inside the solenoid;

B = μ₀ × N × [tex]I[/tex]  

we substitute

B = ( 4π × 10⁻⁷ Tm/A ) × 4000 × 0.235  

B = 1.1881 × 10⁻³ T

Now, Initial flux through the coil is;

∅₁ = NBA = NBπr²  

and the final flux  

∅₂ = 0        

so, the εmf induced ε = -Δ∅/Δt = -( ∅₂ - ∅₁ ) / Δt

= -( 0 - NBπr² ) / Δt  

= NBπr² / Δt    

a)

for N = 7

ε = [ 7 × ( 1.1881 × 10⁻³ ) × π( 1.5 × 10⁻² )² ] / 0.40

ε = 14.7 × 10⁻⁶ v

ε = 14.7 μv      

b)

for N = 10

ε = [ 10 × ( 1.1881 × 10⁻³ ) × π( 1.5 × 10⁻² )² ] / 0.40

ε = 21 × 10⁻⁶ v

ε = 21 μv  

 

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