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2 years ago

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A planar loop consisting of four turns of wire, each of which encloses 200 cm2, is oriented perpendicularly to a magnetic field

that increases uniformly in magnitude from 10 mT to 25 mT in a time of 5.0 ms. What is the resulting induced current in the coil if the resistance of the coil is 5.0 ?

1 answer:

spayn [35]2 years ago

7 0

Answer:

0.6A

Explanation:

Area of loop =200cm2 =200 x10 ∧-4m∧2 Change in Magnetic field (B)= 25mT -10mT =15mT time =5ms

From Faraday' s law of induction EMF(E)= change in magnetic field/time

E= 15mT/5ms

Note, that one weber per second is equivalent to one volt.

= 3V

from Ohm's law I =E/R

=3/5 =0.6A

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Someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (The block of wood is sitting on a frictionless sur

d1i1m1o1n [39]

Answer:

The speed of bullet and wooden bock coupled together, V = 22.22 m/s

Explanation:

Given that,

Mass of the bullet, m = 0.04 Kg

Mass of the wooden block, M = 0.5 Kg

The initial velocity of the bullet, u = 300 m/s

The initial velocity of the wooden block, U = 0 m/s

The final velocity of the bullet and wooden bock coupled together, V = 0 m/s

According to the conservation of linear momentum, the total momentum of the body after impact is equal to the total momentum before impact.

Therefore,

mV + MV = mu + MU

V(m+M) = mu

V = mu/(m+M)

Substituting the values in the above equation,

V = 0.04 Kg x 300 m/s / (0.04 Kg+ 0.5 Kg)

= 22.22 m/s

Hence, the speed of bullet and wooden bock coupled together, V = 22.22 m/s

8 0

2 years ago

a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres

shusha [124]

Answer:

Part a) When collision is perfectly inelastic

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b) When collision is perfectly elastic

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

$mv_m = (m + M)v$

so we have

$v = \frac{mv_m}{m + M}$

now we know that in order to complete the circle we will have

$v = \sqrt{5Rg}$

$\frac{mv_m}{m + M} = \sqrt{5Rg}$

now we have

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b)

Now we know that collision is perfectly elastic

so we will have

$v = \frac{2mv_m}{m + M}$

now we have

$\sqrt{5Rg} = \frac{2mv_m}{m + M}$

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

6 0

2 years ago

A bee wants to fly to a flower located due North of the hive on a windy day. The wind blows from East to West at speed 6.68 m/s.

Aleonysh [2.5K]

Answer: $53.31\°$ East of North

Explanation:

We have the following data:

Speed of the wind from East to West: $6.68 m/s$

Speed of the bee relative to the air: $8.33 m/s$

If we graph these speeds (which in fact are velocities because are vectors) in a vector diagram, we will have a right triangle in which the airspeed of the bee (its speed relative to te air) is the hypotense and the two sides of the triangle will be the <u>Speed of the wind from East to West</u> (in the horintal part) and the <u>speed due North relative to the ground</u> (in the vertical part).

Now, we need to find the direction the bee should fly directly to the flower (due North):

$sin \theta=\frac{Windspeed-from-East-to-West}{Speed-bee-relative-to-air}$

$sin \theta=\frac{6.68 m/s}{8.33 m/s}$

Clearing $\theta$ :

$\theta=sin^{-1} (\frac{6.68 m/s}{8.33 m/s})$

$\theta=53.31\°$

6 0

2 years ago

A farmer is using a rope and pulley to lift a bucket of water from the bottom of a well. the farmer uses a force f1=57.5 n to pu

AlladinOne [14]

The first problem cannot be solve because you did give the distance or length of the rope, because work = distance x force. i can only solve the the second problem. since the bucket is moving up then force due to gravity is going down, then the net force is:
Fnet = F1 - Fg
where Fg = mg
g is the accelaration due to gravity ( 9.81 m/s^2)
Fnet = 57.5 N - (3.9 kg)(9.81) N
Fnet = 19.24 N

4 0

2 years ago

A book is pushed with an initial horizontal velocity of 5.0 meters per second off the top of a 1.19 meter high desk. How far awa

kipiarov [429]

Answer:

2.45 m

Explanation:

First of all, we have to calculate the time of flight of the book, by using the equation for the vertical motion:

$h=\frac{1}{2}gt^2$

where

h = 1.19 m is the vertical distance covered by the book

g = 9.8 m/s^2 is the acceleration of gravity

t is the time of flight

Solving for t,

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(1.19)}{9.8}}=0.49 s$

Now we can find the horizontal distance covered by the book, which is given by

$d=v_x t$

where

$v_x = 5.0 m/s$ is the horizontal velocity

t = 0.49 s is the time of flight

Substituting,

$d=(5.0)(0.49)=2.45 m$

So the book lands 2.45 m away.

8 0

2 years ago