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leva [86]
3 years ago
15

You have a circular loop carrying wire in a large uniform magnetic field because it's a current-carrying loop in a magnetic fiel

d, there will be a torque acting on it. Your friend wants to increase the torque by doubling the loop over itself so that it now makes two circular loops on top of each other by using the same length of wire.
What is the new torque in terms of the old torque?
Physics
1 answer:
Mama L [17]3 years ago
5 0

Answer:

The torque experienced by the new loop is 1/4 of that experienced by the old loop.

Explanation:

The torque τ = IBASinΦ

Assuming the direction of current flow through the loop is perpendicular to that of the magnetic field, SinΦ = Sin90° = 1 and

τ = IBA. With the current flowing through the loop and the magnetic field remaining the same for both cases, then

τ ∝ A. The area is equal to πD² where D is the diameter so

τ ∝ D². Let the old loop have a diameter D. Using the same length of wire, the act of doubling the loop (making 2 turns of wire) reduces the diameter by a factor of 1/2. So the new diameter is D/2

τ ∝ (D/2)² which means that τ ∝ D²/4. So the new torque is 1/4 the old torque.

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1. Consider three objects: Object A is a hoop of mass m and radius r; Object B is a sphere
Zarrin [17]

a) Object C (the disk) has the greatest rotational inertia (\frac{3}{2}mr^2)

b) Object B (the sphere) has the smallest rotational inertia (\frac{4}{5}mr^2)

Explanation:

The moments of inertia of the three objects are the following:

1) For a hoop of negligible thickness, it is

I=MR^2

where M is its mass and R its radius. For the hoop in this problem,

M = m

R = r

Therefore, its moment of inertia is

I=(m)(r)^2=mr^2

2) For a solid sphere, the moment of inertia is

I=\frac{2}{5}MR^2

where M is its mass and R its radius. For the sphere in this problem,

M = 2m

R = r

Therefore, its moment of inertia is

I=\frac{2}{5}(2m)(r)^2=\frac{4}{5}mr^2

3) For a disk of negligible thickness, the moment of inertia is

I=\frac{1}{2}MR^2

where M is its mass and R its radius. For the disk in this problem,

M = 3m

R = r

Therefore, its moment of inertia is

I=\frac{1}{2}(3m)(r)^2=\frac{3}{2}mr^2

So now we can answer the two questions:

a) Object C (the disk) has the greatest rotational inertia (\frac{3}{2}mr^2)

b) Object B (the sphere) has the smallest rotational inertia (\frac{4}{5}mr^2)

Learn more about inertia:

brainly.com/question/2286502

brainly.com/question/691705

#LearnwithBrainly

7 0
4 years ago
If you know the answer please answer the following question down in the picture below.
Harlamova29_29 [7]
I think it’s the second one: magnetic domains must come in pairs—one north and one south
3 0
4 years ago
A skateboarder rolls off a 2.5 m high bridge into the river. If the skateboarder was originally moving at 7.0 m/s, how much time
saul85 [17]

Answer:

  t = 0.714 s and  x = 5.0 m

Explanation:

This is a projectile throwing exercise, in this case when the skater leaves the bridge he goes with horizontal speed

         vₓ = 7.0 m / s

Let's find the time it takes to get to the river

         y = y₀ + v_{oy} t - ½ g t²

the initial vertical speed is zero and when it reaches the river its height is zero

        0 = y₀ + 0 - ½ g t²

        t = \sqrt{\frac{2y_o}{g}  }

        t = ra 2 2.5 / 9.8

        t = 0.714 s

the distance traveled is

       x = vₓ t

       x = 7.0 0.714

       x = 5.0 m

3 0
3 years ago
A 45.0-kg sample of ice is at 0.00°C. How much heat is needed to melt it? For water, Lf=334 kJ/kg and Lv=2257 kJ/kg 
Aleonysh [2.5K]

Heat required to change the phase of ice is given by

Q = m* L

here

m = mass of ice

L = latent heat of fusion

now we have

m = 45 kg

L = 334 KJ/kg

now by using above formula

Q = 45 * 334 * 10^3

Q = 1.5 * 10^7 J

In KJ we can convert this as

Q = 1.5 * 10^4 kJ

so the correct answer is D option

7 0
3 years ago
Read 2 more answers
According to Kepler's Second Law the radius vector drawn from the Sun to a planet Multiple Choice is the same for all planets. s
Mademuasel [1]

Answer:

sweeps out equal areas in equal times.

Explanation:

As we know that there is no torque due to Sun on the planets revolving about the sun

so we will have

\tau_{net} = 0

now we have

\frac{dL}{dt}= 0

now we also know that

Area = \frac{1}{2}r^2d\theta

so rate of change in area is given as

\frac{dA}{dt} = \frac{1}{2}r^2\frac{d\theta}{dt}

so we will have

\frac{dA}{dt} = \frac{1}{2}r^2\omega

\frac{dA}{dt} = \frac{L}{2m}

since angular momentum and mass is constant here so

all planets sweeps out equal areas in equal times.

4 0
4 years ago
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