Differences between angle of twist and angle of shear

Answers

Answer 1

Answer:

idek

Explanation:


Related Questions

what is the distance time how can we find the speed of an object from its distance time graph​

Answers

Answer:

speed is the gradient of the graph

Answer:

Speed is the slope of a distance time graph.

Explanation:

Speed= d/t

Slope is equal to rise/run

If the rise of the graph is the distance and the run is the time, calculating slope is the equivalent of calculating average speed.

The New England Merchants Bank Building in Boston is 152 m high. On windy days it sways with a frequency of 0.18 Hz , and the acceleration of the top of the building can reach 1.9 % of the free-fall acceleration, enough to cause discomfort for occupants.

Required:
What is the total distance, side to side, that the top of the building moves during such an oscillation?

Answers

Answer:

The total distance, side to side, that the top of the building moves during such an oscillation is approximately 0.291 meters.

Explanation:

Let suppose that the building is experimenting a Simple Harmonic Motion due to the action of wind. First, we determine the angular frequency of the system ([tex]\omega[/tex]), in radians per second:

[tex]\omega = 2\pi\cdot f[/tex] (1)

Where [tex]f[/tex] is the frequency, in hertz.

If we know that [tex]f = 0.18\,hz[/tex], then the angular frequency of the system is:

[tex]\omega = 2\pi\cdot (0.18\,hz)[/tex]

[tex]\omega \approx 1.131\,\frac{rad}{s}[/tex]

The maximum acceleration experimented by the system is represented by the following formula, of which we estimate amplitude of the oscillation:

[tex]r\cdot g = \omega^{2}\cdot A[/tex] (2)

Where:

[tex]r[/tex] - Ratio of real acceleration to free-fall acceleration, no unit.

[tex]g[/tex] - Free-fall acceleration, in meters per square second.

[tex]A[/tex] - Amplitude, in meters.

If we know that [tex]\omega \approx 1.131\,\frac{rad}{s}[/tex], [tex]r = 0.019[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the amplitude of the oscillation is:

[tex]A = \frac{r\cdot g}{\omega^{2}}[/tex]

[tex]A = \frac{(0.019)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{\left(1.131\,\frac{rad}{s} \right)^{2}}[/tex]

[tex]A \approx 0.146\,m[/tex]

The total distance, side to side, is twice the amplitude, that is to say, a value of approximately 0.291 meters.

An electron is moving at speed of 6.3 x 10^4 m/s in a circular path of radius of 1.7 cm inside a solenoid the magnetic field of the solenoid is perpendicular to the plane of the electron's path. Find its relevatn motion.

Answers

Answer:

Here, m=9×10

−31

kg,

q=1.6×10

−19

C,v=3×10

7

ms

−1

,

b=6×10

−4

T

r=

qB

mv

=

(1.6×10

−19

)(6×10

−4

)

(9×10

−31

)×(3×10

7

)

=0.28m

v=

2πr

v

=

2πm

Bq

=

2×(22/7)×9×10

−31

(6×10

−4

)×(1.6×10

−19

)

=1.7×10

7

Hz

Ek=

2

1

mv

2

=

2

1

×(9×10

−31

)×(3×10

7

)

2

J

=40.5×10

−17

J=

1.6×10

−16

40.5×10

−17

keV

=2.53keV

A 40-turn coil has a diameter of 11 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.40 T so that the face of the coil and the magnetic field are perpendicular. Find the magnitude of the emf induced in the coil (in V) if the magnetic field is reduced to zero uniformly in the following times.
(a) 0.30 s V
(b) 3.0 s V
(c) 65 s V

Answers

Answer:

(a) emf = 0.507 V

(b) emf = 0.0507 V

(c) emf = 0.00234 V

Explanation:

Given;

number of turns of the coil, N = 40 turns

diameter of the coil, d = 11 cm

radius of the coil, r = 5.5 cm = 0.055 m

magnitude of the magnetic field, B = 0.4 T

The magnitude of the induced emf is calculated as;

[tex]emf = - N\frac{d\phi}{dt} \\\\where;\\\\\phi \ is \ magnetic \ flux= BA \\\\A \ is the \ area \ of \ the \ coil = \pi r^2 = \pi (0.055)^2 = 0.0095 \ m^2\\\\emf = - N \frac{dB.A}{dt} = -NA\frac{dB}{dt} \\\\emf = -NA\frac{(B_2 - B_1)}{t} \\\\emf = NA \frac{(B_1 - B_2)}{t} \\\\the \ final \ magnetic \ field \ is \ reduced \ to \ zero;\ B_2 = 0\\\\emf = \frac{NAB_1}{t}[/tex]

(a) when the time, t = 0.3 s

[tex]emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{0.3} = 0.507 \ V[/tex]

(b) when the time, t = 3.0 s

[tex]emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{3} = 0.0507 \ V[/tex]

(c) when the time, t = 65 s

[tex]emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{65} = 0.00234 \ V[/tex]

A space ship has four thrusters positioned on the top and bottom, and left and right as shown below. The thrusters can be operated independently or together to help the ship navigate in all directions.
Initially, the Space Probe is floating towards the East, as shown below, with a velocity, v. The pilot then turns on thruster #2.

Select one:

a.
Space ship will have a velocity to the West and will be speeding up.

b.
Space ship will have a velocity to the East and will be speeding up.

c.
Space ship will have a velocity to the East and will be slowing down.

d.
Space ship will have a velocity to the West and will be slowing down.

e.
Ship experiences no change in motion.

Answers

Answer:

The correct answer is - c.  Spaceship will have a velocity to the East and will be slowing down.

Explanation:

In this case, if turned on thruster #2 then it will exert force on the west side as thruster 2 is on the east side and it can be understood by Newton's third law that says each action has the same but opposite reaction.

As the spaceship engine applies force on the east side then according to the law the exhauster gas applies on towards west direction. It will try to decrease the velocity of the spaceship however, the direction of floating still be east side initally.

Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon's tongue accelerates at a remarkable 220 m/s^2 for 20 msms, then travels at constant speed for another 30 ms.

Required:
During this total time of 50 ms, 1/20 of a second, how far does the tongue reach?

Answers

Solution :

We know,

Distance,

[tex]$S=ut+\frac{1}{2}at^2$[/tex]

[tex]$S=ut+0.5(a)(t)^2$[/tex]

For the first 20 ms,

[tex]$S=0+0.5(220)(0.020)^2$[/tex]

S = 0.044 m

In the remaining 30 ms, it has constant velocity.

[tex]$v=u+at$[/tex]

[tex]$v=0+(220)(0.020)[/tex]

v = 4.4 m/s

Therefore,

[tex]$S=ut+0.5(a)(t)^2$[/tex]

[tex]$S'=4.4 \times 0.030[/tex]

S' = 0.132 m

So, the required distance is = S + S'

                                              = 0.044 + 0.132

                                              = 0.176 m

Therefore, the tongue can reach = 0.176 m or 17.6 cm

Answer:

The total distance is 0.176 m.

Explanation:

For t = 0 s to t = 20 ms

initial velocity, u = 0

acceleration, a = 220 m/s^2

time, t = 20 ms

Let the final speed is v.

Use first equation of motion

v = u + at

v = 0 + 220 x 0.02 = 4.4 m/s

Let the distance is s.

Use second equation of motion

[tex]s = u t + 0.5 at^2\\\\s = 0 + 0.5 \times 220 \times 0.02\times 0.02\\\\s = 0.044 m[/tex]

Now the distance is

s' = v x t

s' = 4.4 x 0.03 = 0.132 m

The total distance is

S = s + s' = 0.044 + 0.132 = 0.176 m

1:
Forces and Motion:Question 2
A car is travelling east, when suddenly a more massive car travelling
north hits it with a greater force. What is likely to happen to the car
that was originally travelling east?

Answers

Explanation:

the car will be brought back

Help please!!!!!!!!! I will mark brainliest!!!

Answers

Answer:

solving for: velocity

equation: velocity = distance / time

substitution: velocity = 1425 km / 12.5 hrs

answer: 114 km/hr

What particles in an atom can increase and decrease in number without changing the identity of the elements

Answers

Answer:

The number of neutrons or electrons in an atom can change without changing the identity of the element.

A 0.160 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.710 m/s. It has a head-on collision with a 0.296 kg glider that is moving to the left with a speed of 2.23 m/s. Suppose the collision is elastic.

Required:
a. Find the magnitude of the final velocity of the 0.157kg glider.
b. Find the magnitude of the final velocity of the 0.306kg glider.

Answers

The masses of the gliders provided in the question differ from the masses mentioned in the "Required" section. I'll use the first masses throughout.

Momentum is conserved, so the total momentum of the system is the same before and after the collision:

m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'

==>

(0.160 kg) (0.710 m/s) + (0.296 kg) (-2.23 m/s) = (0.160 kg) v₁' + (0.296 kg) v₂'

==>

-0.546 kg•m/s ≈ (0.160 kg) v₁' + (0.296 kg) v₂'

where v₁' and v₂' are the gliders' respective final velocities. Notice that we take rightward to be positive and leftward to be negative.

Kinetic energy is also conserved, so that

1/2 m₁ v₁² + 1/2 m₂ v₂² = 1/2 m₁ (v₁' + 1/2 m₂ (v₂'

or

m₁ v₁² + m₂ v₂² = m₁ (v₁' + m₂ (v₂'

==>

(0.160 kg) (0.710 m/s)² + (0.296 kg) (-2.23 m/s)² = (0.160 kg) (v₁' + (0.296 kg) (v₂'

==>

1.55 kg•m²/s² ≈ (0.160 kg) (v₁' + (0.296 kg) (v₂'

Solve for v₁' and v₂'. Using a calculator, you would find two solutions, one of which we throw out because it corresponds exactly to the initial velocities. The desired solution is

v₁' ≈ -3.11 m/s

v₂' ≈ -0.167 m/s

and take the absolute values to get the magnitudes.

If you want to instead use the masses from the "Required" section, you would end up with

v₁' ≈ -3.18 m/s

v₂' ≈ -0.236 m/s

A object of mass 3.00 kg is subject to a force Fx that varies with position as in the figure below. A coordinate plane has a horizontal axis labeled x (m) and a vertical axis labeled Fx (N). There are three line segments. The first segment runs from the origin to (4,3). The second segment runs from (4,3) to (11,3). The third segment runs from (11,3) to (17,0). (a) Find the work done by the force on the object as it moves from x = 0 to x = 4.00 m. J (b) Find the work done by the force on the object as it moves from x = 4.00 m to x = 11.0 m. J (c) Find the work done by the force on the object as it moves from x = 11.0 m to x = 17.0 m. J (d) If the object has a speed of 0.450 m/s at x = 0, find its speed at x = 4.00 m and its speed at x = 17.0 m.

Answers

Answer:

Explanation:

An impulse results in a change of momentum.

The impulse is the product of a force and a distance. This will be represented by the area under the curve

a) W = ½(4.00)(3.00) = 6.00 J

b) W = (11.0 - 4.00)(3.00) = 21.0 J

c) W = ½(17.0 - 11.0)(3.00) = 9.00 J

d) ASSUMING the speed at x = 0 is in the direction of applied force

½(3.00)(v₄²) = ½(3.00)(0.450²) + 6.00

v₄ = 2.05 m/s

½(3.00)(v₁₇²) = ½(3.00)(0.450²) + 6.00 + 21.0 + 9.00

v₁₇ = 4.92 m/s

If the initial speed is NOT in the direction of applied force, the final speed will be slightly less in both cases.

Good evening everyone Help me i n my hw ,The wall of cinema hall are covered with sound absorbing materials. Why?Answer it ASAP.Good day ​

Answers

what do you mean about it

Convert the unit of 0.00023 kilograms into grams. (Answer in scientific notation)

Answers

Answer:

2.3  ×  [tex]10^{-1}[/tex]  

Explanation:

1 kg = 1000 g.

0.00023 kg x 1000 g = 0.23 grams

Answer:

0.23×10⁴

Explanation:

kilogram to gram ÷ 1000

0.00023kg ÷ 1000

=0.23g

scientific notation=0.23×10⁴

There are 5640 lines per centimeter in a grating that is used with light whose wavelegth is 455 nm. A flat observation screen is located 0.661 m from the grating. What is the minimum width that the screen must have so the centers of all the principal maxima formed on either side of the central maximum fall on the screen

Answers

The minimum width of the screen is 34 cm.

For a diffraction grating, dsinθ = mλ where d = grating spacing = 1/5640 lines per cm = 1/5640 cm per line = 1/5640 × 10⁻² m per line, θ = angle between principal maximum and the center axis of the grating, m = order of maxima = 1 (since we require the position of the principal maximum) and λ = wavelength = 455 nm = 455 × 10⁻⁹ m

So, sinθ = mλ/d

Also tanθ = L/D where θ = angle between principal maximum and the center axis of the grating, L = distance between central maximum and principal maximum and D = distance between grating and screen = 0.661 m.

For small angles sinθ ≈ tanθ

So, mλ/d = L/D

making L subject of the formula, we have

L = mλD/d

L = 1 × 455 × 10⁻⁹ m × 0.661 m ÷  1/5640 × 10⁻² m per line

L = 1 × 455 × 10⁻⁹ m × 0.661 m  × 5640 × 10² line per m

L = 1696258.2 × 10⁻⁷ m

L = 0.16963 m

L ≅ 0.17 m

So, for centers of all the principal maxima formed on either side of the central maximum fall on the screen, the minimum width of the screen is w = 2L.

So, w = 2 × 0.17 m

w = 0.34 m

w = 34 cm

So for the centers of all the principal maxima formed on either side of the central maximum fall on the screen, the minimum width of the screen is 34 cm.

Learn more about diffraction grating here:

https://brainly.com/question/15712101

a point object is 10 cm away from a plane mirror while the eye of an observer(pupil diameter is 5.0 mm) is 28 cm a way assuming both eye and the point to be on the same line perpendicular to the surface find the area of the mirror used in observing the reflection of the point

Answers

Answer:

1.37 mm²

Explanation:

From the image attached below:

Let's take a look at the two rays r and r' hitting the same mirror from two different positions.

Let x be the distance between these rays.

[tex]d_o =[/tex] distance between object as well as the mirror

[tex]d_{eye}[/tex] = distance between mirror as well as the eye

Thus, the formula for determining the distance between these rays can be expressed as:

[tex]x = 2d_o tan \theta[/tex]

where; the distance between the eye of the observer and the image is:

[tex]s = d_o + d_{eye}[/tex]

Then, the tangent of the angle θ is:

[tex]tan \theta = \dfrac{R}{d_o+d_{eye}}[/tex]

replacing [tex]tan \theta = \dfrac{R}{d_o+d_{eye}}[/tex] into [tex]x = 2d_o tan \theta[/tex], we have:

[tex]x = 2d_o \Big( \dfrac{R}{d_o+d_{eye}}\Big)[/tex]

[tex]x = 2(10) \Big( \dfrac{0.25}{10+28}\Big)[/tex]

[tex]x = 20\Big( \dfrac{0.25}{38}\Big) cm[/tex]

x = (0.13157 × 10) mm

x = 1.32 mm

Finally, the area A = π r²

[tex]A = \pi(\frac{x}{2})^2[/tex]

[tex]A = \pi(\frac{1.32}{2})^2[/tex]

A = 1.37 mm²

Cuando el pistón tiene un volumen de 2x10^-4 m^3, el gas en el pistón está a una presión de 150 kPa. El área del pistón es 0.00133 m^2. Calcular la fuerza que el gas ejerce sobre el embolo del pistón.

Answers

Answer:

F = 1.128 10⁸ Pa

Explanation:

Pressure is defined by

         P = F / A

If the gas is ideal for equal force eds on all the walls, so on the piston area we have

        F = P A

We reduce the pressure to the SI system

       P = 150 kpa (1000 Pa / 1kPa = 150 103 Pa

we calculate

       F = 150 10³ / 0.00133

       F = 1.128 10⁸ Pa

If ∆H = + VE , THEN WHAT REACTION IT IS
1) exothermic
2) endothermic​

Answers

Answer:

endothermic

Explanation:

An endothermic is any process with an increase in the enthalpy H (or internal energy U) of the system. In such a process, a closed system usually absorbs thermal energy from its surroundings, which is heat transfer into the system.

A spring whose stiffness is 3500 N/m is used to launch a 4 kg block straight up in the classroom. The spring is initially compressed 0.2 m, and the block is initially at rest when it is released. When the block is 1.3 m above its starting position, what is its speed

Answers

Answer:

the speed of the block at the given position is 21.33 m/s.

Explanation:

Given;

spring constant, k = 3500 N/m

mass of the block, m = 4 kg

extension of the spring, x = 0.2 m

initial velocity of the block, u = 0

displacement of the block, d =1.3 m

The force applied to the block by the spring is calculated as;

F = ma = kx

where;

a is the acceleration of the block

[tex]a = \frac{kx}{m} \\\\a = \frac{(3500) \times (0.2)}{4} \\\\a = 175 \ m/s^2[/tex]

The final velocity of the block at 1.3 m is calculated as;

v² = u² + 2ad

v² = 0 + 2ad

v² = 2ad

v = √2ad

v = √(2 x 175 x 1.3)

v = 21.33 m/s

Therefore, the speed of the block at the given position is 21.33 m/s.

The speed of the block at a height of 1.3 m above the starting position is 21.33 m/s

To solve this question, we'll begin by calculating the acceleration of the block.

How to determine the acceleration Spring constant (K) = 3500 N/m Mass (m) = 4 KgCompression (e) = 0.2 mAcceleration (a) =?

F = Ke

Also,

F = ma

Thus,

ma = Ke

Divide both side by m

a = Ke / m

a = (3500 × 0.2) / 4

a = 175 m/s²

How to determine the speed Initial velocity (u) = 0 m/sAcceleration (a) = 175 m/s²Distance (s) = 1.3 mFinal velocity (v) =?

v² = u² + 2as

v² = 0² + (2 × 175 × 1.3)

v² = 455

Take the square root of both side

v = √455

v = 21.33 m/s

Learn more about spring constant:

https://brainly.com/question/9199238

19 point please please answer right need help

Block on an incline
A block of mass m1 = 3.9 kg on a smooth inclined plane of angle 38is connected by a cord over a small frictionless
pulley to a second block of mass m2 = 2.6 kg hanging vertically. Take the positive direction up the incline and use 9.81
m/s2 for g.
What is the tension in the cord to the nearest whole number?

Answers

Explanation:

We can write Newton's 2nd law as applied to the sliding mass [tex]m_1[/tex] as

[tex]T - m_1g\sin38 = m_1a\:\:\:\:\:\:\:(1)[/tex]

For the hanging mass [tex]m_2,[/tex] we can write NSL as

[tex]T - m_2g = -m_2a\:\:\:\:\:\:\:(2)[/tex]

We need to solve for a first before we can solve the tension T. So combining Eqns(1) & (2), we get

[tex](m_1 + m_2)a = m_2g - m_1g\sin38[/tex]

or

[tex]a = \left(\dfrac{m_2 - m_1\sin38}{m_1 + m_2}\right)g[/tex]

[tex]\:\:\:\:= 0.30\:\text{m/s}^2[/tex]

Using this value for the acceleration on Eqn(2), we find that the tension T is

[tex]T = m_2(g - a) = (2.6\:\text{kg})(9.51\:\text{m/s}^2)[/tex]

[tex]\:\:\:\:=24.7\:\text{N}[/tex]

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

S.I unit for distance =______

(A) m (B)cm

(c) km (d) mm

Answers

Answer:

opinion a

Explanation:

the si units of distance is metre (m)

Answer:

A

Explanation:

A capacitor consists of a set of two parallel plates of area A separated by a distance d. This capacitor is connected to a battery and charged until its plates carry charges If the separation between the plates is doubled, the electrical energy stored in the capacitor will

Answers

Answer:

The electrical energy stored in the capacitor will be cut in half.

Explanation:

The energy in a capacitator is given by E=C[tex]V^{2}[/tex]/2 and the formula for the Capacitance in a capacitator is C= [tex]\frac{Q}{V}[/tex] = ε[tex]\frac{A}{d}[/tex] .

So if we replace C = ε[tex]\frac{A}{d}[/tex]  in the first equation we have:

E = ε[tex]\frac{AV^{2} }{2d}[/tex]

1. Why do only some people get addicted to
drugs?

Answers

Answer:

When drugs are taken in are body are brain release dopamine: which make us feel so pleasure and good, and for this some people are addicted to drugs which makes them feel good. on other hand damaging their health.

a vector starts at the point (0.0) and ends at (2,-7) what is the magnitude of the displacement

Answers

Answer:

|x| = √53

Explanation:

We are told that the vector starts at the point (0.0) and ends at (2,-7) .

Thus, magnitude of displacement is;

|x| = √(((-7) - 0)² + (2 - 0)²)

|x| = √(49 + 4)

|x| = √53

Which of the following scientists won a Nobel Prize for pioneering work in the
study of the evolution of stars?
A. Christian Doppler
B. Warren Washington
C. Charles Kuen Kao
-D. Subrahmanyan Chandrasekhar

Answers

Answer:

Subrahmanyan Chandrasekhar

Answer:

D. Subrahmanyan Chandrasekhar

Explanation:

Match the following properties to the type of wave.

Answers

Answer:

hi there

Explanation:

1 - III

2- 1

3-1

HOPE IT HELP YOU

PLz mark me as a BRAINLIST

Explanation:

1 . 3

2. 1

3. 2

I hope it is helpful to you.

A car is running with the velocity of 72 km per hour what will be its velocity after 5 seconds if its acceleration is -2 metre per second square​

Answers

Answer:

initial velocity (u)=72×1000/60×60

=72000/3600

=20m/s

final velocity(v)=v

Time(t)=5s

acceleration(a)=-2m/s

now,

acceleration(a)=v-u/t

-2=v-20/5

-2×5=v-20

-10=v-20

-10+20=v

v=10m/s

The primary purpose of a switch in a circuit is to ___________.

A)either open or close a conductive path
B)change a circuit from parallel to series
C)change a circuit from series to parallel
D)store a charge for later use

Answers

Answer:

store a charge for later use

If a jet travels 350 m/s, how far will it travel each second?

Answers

Answer:

It will travel 350 meters each second.

Explanation:

The unit rate, 350 m/s, tells us that the jet will travel 350 meters per every second elapsed.

Answer:

5.83 seconds

Explanation:

60 seconds in 1 minute

350 meters per second

350/60

=5.83

The armature of an AC generator has 200 turns, which are rectangular loops measuring 5 cm by 10 cm. The generator has a sinusoidal voltage output with an amplitude of 18 V. If the magnetic field of the generator is 300 mT, with what frequency does the armature turn

Answers

Answer:

[tex]f=9.55Hz[/tex]

Explanation:

From the question we are told that:

Number of Turns [tex]N=200[/tex]

Length [tex]l=5cm to 10cm[/tex]

Voltage [tex]V=18V[/tex]

Magnetic field [tex]B=300mT[/tex]

Generally, the equation for Frequncy of an amarture is mathematically given by

[tex]f =\frac{ V}{(N B A * 2 pi )}[/tex]

[tex]f =\frac{ 18}{(200 300*10^{-3} (10*10^-2)(5*10^{-2}) * 2 *3.142 )}[/tex]

[tex]f=9.55Hz[/tex]

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