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7nadin3 [17]
3 years ago
9

A single circular loop of wire of radius 0.75 m carries a constant current of 3.0 A. The loop may be rotated about an axis that

passes through the center and lies in the plane of the loop. When the orientation of the normal to the loop with respect to the direction of the magnetic field is 25°, the torque on the coil is 1.8 N ⋅ m. What is the magnitude of the uniform magnetic field exerting this torque on the loop?
Physics
1 answer:
NISA [10]3 years ago
6 0

Answer:

B = 0.8 T

Explanation:

It is given that,

Radius of circular loop, r = 0.75 m

Current in the loop, I = 3 A

The loop may be rotated about an axis that passes through the center and lies in the plane of the loop.

When the orientation of the normal to the loop with respect to the direction of the magnetic field is 25°, the torque on the coil is 1.8 Nm.

We need to find the magnitude of the uniform magnetic field exerting this torque on the loop. Torque acting on the loop is given by :

\tau=NIAB\sin\theta

B is magnetic field

B=\dfrac{\tau}{NIA\sin\theta}\\\\B=\dfrac{1.8}{1\times \pi \times (0.75)^2\times 3\times \sin(25)}\\\\B=0.8\ T

So, the magnitude of the uniform magnetic field exerting this torque on the loop is 0.8 T.

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Answer:gjv

Explanation:

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You have probably noticed that carrying a person in a pool of water is much easier than carrying a person through air. To unders
Kruka [31]

Answer:

The answer is "0.91238 and 744.8"

Explanation:

In this scenario it is easier to take a person to the water-pool than to transport the people in the air, as the person's strength is increased by water upwards:

f_b \to m \to mg =person \\\\F_B \ in\  air = v\ & air\  g \\\\

               =0.076 \times 1.225 \times 9.8 \\\\ =0.91238 \ N\\\\

F_B \ in \ water = v  \& water \ g \\\\

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6 0
2 years ago
The wing of an airplane experiences the forces as depicted in the vector diagram to the right. Using both one and two dimensiona
Vedmedyk [2.9K]

Answer:

A.) 3605.6 N

B.) 33.7 degree

Explanation:

To find the result force acting on the wing of the airplane, we need to resolve the forces into x and y components

Resolving into x component :

Sum of forces = 3500 - 500 = 3000N

Resolving into y component:

Sum of forces = 2000N

Resultant force Fr = sqrt ( Fx^2 + Fy^2)

Fr = sqrt ( 3000^2 + 2000^2 )

Fr = sqrt ( 9000000 + 4000000 )

Fr = sqrt ( 13000000)

Fr = 3605.6 N

Therefore, resultant force acting on the wing is 3605.6 N

The direction of the vector will be:

Tan Ø = Fy / Fx

Substitute Fx and Fy into the formula

Tan Ø = 2000 / 3000

Tan Ø = 0.66666

Ø = tan^-1(0. 66666)

Ø = 33.7 degree.

6 0
3 years ago
A brick is resting on a smooth wooden board that is at a 30° angle. What is one way to overcome the static friction that is hold
lubasha [3.4K]

Answer:

to overcome the out of friction we must increase the angle of the plane

Explanation:

To answer this exercise, let's propose the solution of the problem, write Newton's second law. We define a coordinate system where the x axis is parallel to the plane and the other axis is perpendicular to the plane.

X axis

       fr - Wₓ = m    a                      (1)

Y axis  

       N- W_{y} = 0

       N = W_{y}

let's use trigonometry to find the components of the weight

        sin θ = Wₓ / W

        cos θ = W_{y} / W

        Wₓ = W sin θ

        W_{y} = W cos θ

the friction force has the formula

         fr = μ N

         fr = μ Wy

         fr = μ mg cos θ

from equation 1

at the point where the force equals the maximum friction force

in this case the block is still still so a = 0

           F = fr

           F = (μ  mg) cos θ

We can see that the quantities in parentheses with constants, so as the angle increases, the applied force must be less.

This is the force that balances the friction force, any force slightly greater than F initiates the movement.

Consequently, to overcome the out of friction we must increase the angle of the plane

the correct answer is to increase the angle of the plane

4 0
3 years ago
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