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3 years ago

A rectangular loop consists of 80 turns, each carrying a current 8

Physics

1 answer:

liberstina [14]3 years ago

3 0

Magnetic moment of loop is given by the formula

$M = NiA$

$M = 80*8 *0.4*0.3$

$M = 76.8$

now magnetic field is given here

$B = 40 * 10^{-3} T$

Now the torque on the loop is given by

$T = MBsin\theta$

$T = 76.8* 40*10^{-3}*sin30$

$T = 1.536 Nm$

so above is the torque on the loop.

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3 years ago

An elevator provides 21 000 w of power during a 12 s ride. how much work does the elevator do?

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3 years ago

the moon revolves around the earth in a nearly circular orbit kept by gravitational force exerted by the earth work done will be

rodikova [14]

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Zero because the applied force is perpendicular to the motion of the object.

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2 years ago

HELP! Can an object that is in motion have balanced forces? Explain your answer

Elan Coil [88]

Answer:

Yes.

Explanation:

Newton's first law says that an object in motion stays in motion and an object at rest stays at rest until acted upon by an unbalanced force.

If an object in motion has balanced forces, it will stay in motion. For example, if an object is falling at terminal velocity (for example, a parachuter), then the force of gravity is equal and opposite to the force of air resistance. The forces are balanced, and the object continues to fall at a constant speed.

6 0

3 years ago

A curve that has a radius of 90 m is banked at an angle of =10.8∘. If a 1100 kg car navigates the curve at 75 km/h without skidd

PilotLPTM [1.2K]

The minimum coefficient of static friction between the pavement and the tires is 0.69.

The given parameters;

radius of the curve, r = 90 m
angle of inclination, θ = 10.8⁰
speed of the car, v = 75 km/h = 20.83 m/s
mass of the car, m = 1100 kg

The normal force on the car is calculated as follows;

$F_n = mgcos(\theta)$

The frictional force between the car and the road is calculated as;

$F_k = \mu_k F_n\\\\F_k = \mu_k mgcos(\theta)$

The net force on the car is calculated as follows;

$mgsin(\theta) + \mu_s mgcos(\theta) = \frac{mv^2}{r} \\\\mg(sin\theta \ + \ \mu_s cos\theta)= \frac{mv^2}{r} \\\\g(sin\theta \ + \ \mu_s cos\theta)= \frac{v^2}{r}\\\\sin\theta \ + \ \mu_s cos\theta = \frac{v^2}{rg}\\\\\mu_s cos\theta = sin\theta \ + \ \frac{v^2}{rg}\\\\\mu_s = \frac{sin\theta}{cos \theta} + \frac{v^2}{cos (\theta)rg}\\\\\mu_s = tan(\theta) + \frac{v^2}{cos (\theta)rg}\\\\\mu_s = tan(10.8) + \frac{(20.83)^2}{cos(10.8) \times 90 \times 9.8} \\\\\mu_s = 0.19 + 0.5\\\\$

$\mu_s = 0.69$

Thus, the minimum coefficient of static friction between the pavement and the tires is 0.69.

Learn more here:brainly.com/question/15415163

8 0

3 years ago