A force is always required to move an object from rest

An AC generator consists of 20 circular loops of wire with an area of 75 cm2. It has a maximum induced voltage of 24 V. If its a

Monica [59]

Faraday's law allows us to find the magnetic field that produces the emf in the rotating system is:

The magnetic field is: B = 0.424 T

Faraday's law of induction states that when the magnetic flux changes in time, an induced electromotive force is produced.

fem = $- \frac{d \Phi_B }{dt}$

where fem is the induced electromotive force and Ф the flux,

The magnetic flux is the scalar product of the field and the area.

$\Phi_B = B . A = B A \ cos \theta$

In this case we have several turns, so the expression remains.

fem = $- N B A \ \frac{d cos \theta}{dt}$

Indicate that the turns rotate at a constant frequency, therefore we can use the uniform rotational motion ratio.

θ = w t

We substitute

$fem = - N B A \ \frac{d \ cos \ wt}{dt}\\fem = N B A w sin \ wt$

the maximum induced electromotive force occurs when the sine function is ±1

fem = N B A w

They indicate that the fem = 24 V, the number of the turn is N = 20, the area is A = 75 cm² = 75 10⁻⁴ m² and the frequency f = 60 Hz

Frequency and angular velocity are related.

w = 2π f

We substitute.

fem = N B A 2π f

$B = \frac{fem }{2 \pi \ NA \ f}$

Let's calculate.

$B= \frac{24 }{2\pi \ 20 \ 75 \ 10^{-4} 60}$ B = 24 / 2pi 20 75 10-4 60

B = 0.424 T

In conclusion, using Faraday's law we can find the magnetic field that produces the emf in the rotating system is:

The magnetic field is; B = 0.424 T

Learn more about Faraday's law here: brainly.com/question/24617581

8 0

3 years ago

A record of travel along a straight path is as follows: 1. Start from rest with constant acceleration of 2.65 m/s2 for 17.0 s. 2

nlexa [21]

Hello

Let's solve the problem in the three different steps

1) Uniformly accelerated motion, with acceleration $a_1 = 2.65~m/s^2$ and for a total time of $t_1=17~s$ . The body is initially at rest, so the distance covered is given by
$S= \frac{1}{2}a_1t_1^2=382.9~m$
Calling $v_f$ and $v_i$ the final and initial velocity, and since the $v_i=0~m/s$ because the body starts from rest, we can use
$a= \frac{v_f-v_i}{t}$
to find the final velocity after this first leg:
$v_{f}=v_i+a_1t_1=45~m/s$
And the average velocity in this first leg is
$v_1= \frac{v_f+v_i}{2}=22.5~m/s$

2) Uniform motion. The velocity is constant and it is equal to the final velocity of the first leg: $v_2=45~m/s$ . This is also the average velocity of the second leg.
The total time of this second leg is $t_2=1.60~min = 96~s$ . The distance covered is given by
$S_2=v_2t_2=45~m/s \cdot 96~s=4320~m$

3) Uniformly decelerated motion, with constant deceleration of $a_3=-9.39~m/s^2$ and for a total time of $t_3=4.8~s$ . Here, the initial velocity of the body is the final velocity of the previous leg, i.e. $v_i=45~m/s$ . Therefore, the distance covered in this leg is given by
$S_3=v_i t_3 + \frac{1}{2} a_3 t^2 =107.8~m$
The final velocity in this leg is given by
$v_f=v_i+at=45~m/s-9.39~m/s^2 \cdot 4.8~s = -0.07~m/s$
The negative sign means that after decelerating, the body has started to go in the opposite direction. Similarly to step 1, the average velocity in this leg is given by
$v_3 = \frac{1}{2}(v_f+v_i)= \frac{1}{2}(-0.07~m/s+45~m/s)= 22.5~m/s$

4) Finally, the total distance covered in the motion is
$S=S_1+S_2+S_3=382.9~m+4320~m+107.8~m=4810.7~m$
To find the average velocity, we must "weigh" the average velocity of each leg for the correspondent time of that leg:
$v_{ave}= \frac{v_1t_1+v_2t_2+v_3t_3}{t_1+t_2+t_3}=40.8~m/s$

8 0

3 years ago

hey pls help with this HDJSJDH im rlly bad at this kinda stuff :,) i'll mark the correct answer brainliest

timama [110]

I am pretty sure it’s A
The cue exerts force onto the white ball which pushes the blue ball into the direction of the hole.

7 0

3 years ago

Read 2 more answers

If the mass of an object increases, the force acting on it, such as gravitational force, also increases

strojnjashka [21]

Its true because I just took the the test and it was right
I hope i helped you got it right

4 0

4 years ago

А

timurjin [86]

Answer:

i hope it will be useful for you

Explanation:

F=5.6×10^-10N

R=93cm=0.93m

let take m1 and m2 =m²

according to newton's law of universal gravitation

F=m1m2/r²

F=m²/r²

now we have to find masses

F×r²=m²

5.6×10^10N×0.93m=m²

5.208×10^-9=m²

taking square root on b.s

√5.208×10^-9=√m²

so the two masses are m1=7.2×10^-5

and m2=7.2×10^-5

8 0

4 years ago