Ask question

Ask question

All categories

Neporo4naja [7]

3 years ago

15

A 149 turn 149 turn circular coil of radius 2.15 cm 2.15 cm is immersed in a uniform magnetic field that is perpendicular to the

plane of the coil. Over an interval of 0.165 s 0.165 s , the magnetic field strength increases from 50.5 mT 50.5 mT to 95.5 mT 95.5 mT . Find the magnitude of the average emf E avg Eavg induced in the coil during this tim

1 answer:

adoni [48]3 years ago

4 0

Answer:

Explanation:

number of turns, N = 149

radius, r = 2.15 cm

Area, A = πr² = 3.14 x 2.15 x 2.15 x 10^-4 = 1.45 x 10^-3 m^2

Change in magnetic field, ΔB = 95.5 - 50.5 = 45 mT = 45 x 10^-3 T

time, Δt = 0.165 second

induced emf

e = N x dФ/dt

where, dФ be the change in flux.

e = N x A x ΔB/Δt

e = 149 x 1.45 x 10^-3 x 45 x 10^-3 / 0.165

e = 0.058 V

You might be interested in

Electric devices use electric _____ to function

Rina8888 [55]

AC or DC

someone use active current or Direct current

6 0

3 years ago

As an adult, what is the best way to remain healthy?

Zigmanuir [339]

1. Eat a balanced diet and regular exercise. This is the best way because eating right and exercising are is more beneficial than just exercising.

4 0

3 years ago

Read 2 more answers

You are working with a team that is designing a new roller coaster-type amusement park ride for a major theme park. You are pres

Dominik [7]

Answer:

<em>The required constant friction force for the last 20 m is 6,862.8 N</em>

Explanation:

<u>Energy Conversion</u>

There are several ways the energy is manifested in our physical reality. Some examples are Kinetic, Elastic, Chemical, Electric, Potential, Thermal, Mechanical, just to mention some.

The energy can be converted from one form to another by changing the conditions the objects behave. The question at hand states some types of energy that properly managed, will make the situation keep under control.

Originally, the m=220 kg car is at (near) rest at the top of a h=101 m tall track. We can assume the only energy present at that moment is the potential gravitational energy:

$E_1=mgh=220\cdot 9.8\cdot 101=217,756\ J$

For the next x1=230 m, a constant friction force Fr1=350 N is applied until it reaches ground level. This means all the potential gravitational energy was converted to speed (kinetic energy K1) and friction (thermal energy W1). Thus

$E_1=K_1+W_1$

We can compute the thermal energy lost during this part of the motion by using the constant friction force and the distance traveled:

$W_1=F_{r1}\cdot x_1=350\cdot 230=80,500\ J$

This means that the kinetic energy that remains when the car reaches ground level is

$K_1=E_1-W_1=217,756\ J-80,500\ J=137,256\ J$

We could calculate the speed at that point but it's not required or necessary. That kinetic energy is what keeps the car moving to its last section of x2=20 m where a final friction force Fr2 will be applied to completely stop it. This means all the kinetic energy will be converted to thermal energy:

$W_2=F_{r2}\cdot x_2=137,256$

Solving for Fr2

$\displaystyle F_{r2}=\frac{137,256}{20}=6,862.8\ N$

The required constant friction force for the last 20 m is 6,862.8 N

3 0

3 years ago

Which of these is a biotic factor? <br> A) topography <br> B) soil <br> C )air <br> D)bacteria

vitfil [10]

D I might be wroung .....i think this so the answer because biotic are all living things and bacteria is living

7 0

3 years ago

The position of a particle moving along the x axis depends on the time according to the equation x = ct² - bt³, where x is in me

Kamila [148]

Answer:

(a) $\frac{m}{s^2}$

(b) $\frac{m}{s^3}$

(c) 1 s

(d) 20 m

(e) 1 m

(f) $0\frac{m}{s}$

(g) $-12\frac{m}{s}$

(h) $-36\frac{m}{s}$

(i) $-72\frac{m}{s}$

(j) $-6\frac{m}{s^2}$

(k) $-18\frac{m}{s^2}$

(l) $-30\frac{m}{s^2}$

(m) $-42\frac{m}{s^2}$

Explanation:

Since <em>x</em> is measured in meters and <em>t</em> in seconds, constants <em>a </em>and <em>b</em> must have units that gives meters when multiplied by square and cubic seconds respectivly, so that would mean $\frac{m}{s^2}$ for <em>a </em>and $\frac{m}{s^3}$ for <em>b</em>.

We can get the velocity <em>v </em>equation by deriving the position with respect to <em>t</em>, which gives:

$v=6*t-6*t^2$

And the acceleration <em>a</em> equation by deriving again:

$a=6-12*t$

Now for getting the maximun position between 0 and 4, we must find to points where the positions first derivate is equal to cero and evaluate those points. That is <em>v=0</em>, which gives

$6*t-6*t^2=0\\6t*(1-t)=0\\t=0 or t=1$

For <em>t = 0</em>,<em> x = 0</em> so the maximun position is archieved at 1 second, which gives <em>x = 1 meter</em>.

For obtaining it's displacement <em>r</em>, we can integrate the velocity from 0 seconds to 4 seconds, which gives the mean value of the position in that interval:

$r=\frac{1}{4}*(2*4^3-3*4^2)m\\r= 20m$

For the remaining questions, we just replace the values of <em>t</em> on the respective equations.

8 0

4 years ago