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3 years ago

A freely suspended bar magnet always shows north and south direction why

Physics

1 answer:

Masja [62]3 years ago

8 0

<h2>Answer:</h2>

The earth behaves as a magnetic dipole. Therefore a freely suspended magnet always points towards in the north-south direction because the north pole of the suspended magnet attracts the south pole of the earth's magnet which is the geographical north pole of the earth.

hope this help!

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As a dilligent physics student, you carry out physics experiments at every opportunity. At this opportunity, you carry a 1.11-m-

soldier1979 [14.2K]

Answer:

The strength of the magnetic field is 0.0842 mT

Explanation:

Given:

Velocity of rod $v = 3.07$ $\frac{m}{s}$

Length of rod $l = 1.11$ m

Induced emf across the rod $\epsilon = 0.287 \times 10^{-3}$ V

According to the faraday's law

We have a special case for moving rod in magnetic field.

Induced emf in moving rod is given by,

$\epsilon = Blv$

Where $B =$ strength of magnetic field

$B = \frac{\epsilon}{lv}$

$B = \frac{0.287 \times 10^{-3} }{3.07 \times 1.11}$

$B = 0.0842 \times 10^{-3}$ T

$B = 0.0842$ mT

Therefore, the strength of the magnetic field is 0.0842 mT

7 0

3 years ago

If light travels at 3.00

Juliette [100K]

Yes light travels at 3.00

5 0

3 years ago

Read 2 more answers

A plane has a speed of 285 km/h west relative to the air. A wind blows 25 km/h east relative to the ground. what are the planes

geniusboy [140]

Answer:

260 km/h

Explanation:

Motion is said to be relative when the motion observed by the observer depends on the location (frame) of the observer.

When two objects are moving in opposite directions, the relative motion is obtained by subtraction.

Hence the plane is moving at (285 - 25) km/h giving 260 km/h

3 0

3 years ago

A small rescue helicopter has 4 blades, each 4m long and 50kg in mass, and it carries a load of 1000kg. a) Find the rotational k

Tomtit [17]

Answer:

A) KE_rot= 5.26 x 10^(5) J

B) KE_trans = 2 x 10^(5)J

Ratio = 0.38

C) Max height = 52.7m

Explanation:

A) We are given;

m = 50kg

r = 4m

ω = 300 revs/min = 300 x 0.1047 = 31.4 rads/s

The formula for rotational kinetic energy is given as;

KE_rot= (1/2)Iω²

Where I is moment of inertia.

Now, I = mr²/3

But since it has 4 blades, thus,

I = 4mr²/3 = 4 x 50 x 4²/3 = 1067 kg. m²

Thus, KE_rot= (1/2)(1067)(31.4)² = 5.26 x 10^(5) J

B) The formula for translational kinetic energy is given by;

KE_trans = (1/2)mv²

We are given velocity as 20 m/s

Thus,

KE_trans = (1/2)(50)(20)² = 2 x 10^(5)J

To compare this to the rotational kinetic energy, we'll take the ratio. Thus; = KE_trans/KE_rot = 2 x 10^(5)J/5.26 x 10^(5)J = 0.38

C) At maximum height,

KE_rot = PE_gravity

Thus, 5.26 x 10^(5)J = mgh

Where m is the load it carries;

5.26 x 10^(5)J = 1000 x 9.8 x h

So, h = [5.26 x 10^(5)]/(9.8 x 1000) = 52.7m

4 0

4 years ago

A 10-cm-long wire is pulled along a u-shaped conducting rail in a perpendicular magnetic field. the total resistance of the wire

Debora [2.8K]

In the above case we can say that power given by external agent to pull the rod must be equal to the power dissipated in the form of heat due to magnetic induction.

Part a)

when we pull the rod with constant speed then power required will be product of force and velocity

here we will have

$P = F.v$