Ask question

Ask question

All categories

3 years ago

7

A coil consists of 180 turns of wire. Each turn is a square of side d=30 cm, and a uniform magnetic field directed perpendicular

to the plane of the coil is turned on. If the field changes linearly from 0 to 1.22 T in 0.06s, what is the magnitude of the induced force in the coil while the field is changing?

1 answer:

alexira [117]3 years ago

8 0

Answer:

Emf induced i equal to 329.4 volt

Explanation:

Note : Here i think we have to find emf induced in the coil

Number of turns in the coil N= 180

Sides of square d = 30 cm = 0.3 m

So area of the square $A=0.3\times 0.3=0.09m^2$

Magnetic field is changes from 0 to 1.22 T

Therefore $dB=1.22-0=1.22T$

Time interval in changing the magnetic field dt = 0.06 sec

Induced emf is given by

$e=-N\frac{d\Phi }{dt}=-NA\frac{dB}{dt}$

$e=-180\times 0.09\times \frac{1.22}{0.06}=329.4volt$

You might be interested in

3. The center of mass (or center of gravity) of a two-particle system is at the origin. One particle

nika2105 [10]

Answer:

B) (-2.0 m, 0.0 m)

Explanation:

Given:

Mass of particle 1 is, $m_1=2.0\ kg$

Mass of particle 2 is, $m_2=3.0\ kg$

Position of center of mass is, $(x_{cm},y_{cm})=(0,0)$

Position of particle 1 is, $(x_1,y_1)=(3.0\ m,0.0\ m)$

Position of particle 2 is, $(x_2,y_2)=(?\ m,?\ m)$

We know that, the x-coordinate of center of mass of two particles is given as:

$x_{cm}=\frac{m_1x_1+m_2x_2}{m_1+m_2}$

Plug in the values given.

$0=\frac{(2.0\times 3)+(3.0\times x_2)}{2.0+3.0}\\\\0=6+3x_2\\\\3x_2=-6\\\\x_2=\frac{-6}{3}=-2.0\ m$

We know that, the y-coordinate of center of mass of two particles is given as:

$y_{cm}=\frac{m_1y_1+m_2y_2}{m_1+m_2}$

Plug in the values given.

$0=\frac{(2.0\times 0)+(3.0\times y_2)}{2.0+3.0}\\\\0=0+3y_2\\\\3y_2=0\\\\y_2=\frac{0}{3}=0.0\ m$

Therefore, the position of particle 2 of mass 3.0 kg is (-2.0 m, 0.0 m).

So, option (B) is correct.

8 0

3 years ago

One watt is wqual to 1kg .m/s³

djverab [1.8K]

1 watt = 1 joule per second = 1 newton meter per second = 1 kg m2 s-3

6 0

3 years ago

Read 2 more answers

WILL GIVE BRAINLIEST!!!!!!!!!!!! 50 PTS!!!!!!!!!!!!!!! 10 SCIENCE QUESTIONS!!!!

Jet001 [13]

1.b 2.c 3.a 4.b 5.d 6.a 7.a 9.idk 10.c hope thes helps have a nice day and God bless

8 0

3 years ago

Un automóvil acelera de 0 a 140 km/h en 9.8 segundos, determina su aceleración

guajiro [1.7K]

Answer:

The acceleration is 14.28 km/h^2

Explanation:

Step one:

Given data

initial speed u= 0 km/h

final speed v= 140km/h

time t= 9.8 seconds

Required

The acceleration of the car

Step two:

From a= v-u/t

substitute

a= 140-0/9.8

a=140/9.8

a=14.28 km/h^2

6 0

3 years ago

A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 85.0 m/s2

sergey [27]

Answer:

Maximum height attained by the model rocket is 2172.87 m

Explanation:

Given,

Initial speed of the model rocket = u = 0
acceleration of the model rocket = $a\ =\ 85.0 m/s^2$
time during the acceleration = t = 2.30 s

We have to consider the whole motion into two parts

In first part the rocket is moving with an acceleration of a = 85.0 $m/s^2$ for the time t = 2.30 s before the fuel abruptly runs out.

Let $s_1$ be the height attained by the rocket during this time intervel,

$s_1\ =\ ut\ +\ \dfrac{1}{2}at^2\\\Rightarrow s_1\ =\ 0\ +\ 0.5\times 85\times 2.30^2\\\Rightarrow s_1\ =\ 224.825\ m$

And Final velocity at that point be v

$\therefore v\ =\ u\ +\ at\\\Rightarrow v\ =\ 0\ +\ 85.0\times 2.3\\\Rightarrow v\ =\ 195.5\ m/s.$

Now, in second part, after reaching the altitude of 224.825 m the fuel abruptly runs out. Therefore rocket is moving upward under the effect of gravitational acceleration,

Let ' $s_2$ ' be the altitude attained by the rocket to reach at the maximum point after the rocket's fuel runs out,

At that insitant,

initial velocity of the rocket = v = 195.5 m/s.

a = $-g\ =\ -9.81\ m/s^2$
Final velocity of the rocket at the maximum altitude = $v_f\ =\ 0$

From the kinematics,

$v^2\ =\ u^2\ +\ 2as\\\Rightarrow 0\ =\ u^2\ -\ 2gs_2\\\Rightarrow s_2\ =\ \dfrac{u^2}{2g}\\\Rightarrow s_2\ =\ \dfrac{195.5^2}{2\times 9.81}\\\Rightarrow s_2\ =\ 1948.02\ m$

Hence the maximum altitude attained by the rocket from the ground is

$s\ =\ s_1\ +\ s_2\ =\ 224.85\ +\ 1948.02\ =\ 2172.87\ m$

6 0

3 years ago