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igor_vitrenko [27]
3 years ago
15

People have proposed driving motors with the earth's magnetic field. This is possible in principle, but the small field means th

at unrealistically large currents are needed to produce noticeable torques. Suppose a 20-cm-diameter loop of wire is oriented for maximum torque in the earth's field.What current would it need to carry in order to experience a very modest 1.0×10−3N⋅m torque?
Physics
1 answer:
NISA [10]3 years ago
3 0

Answer:

Explanation:

Torque on a loop in a magnetic field

Maximum torque = M B

M is magnetic moment  of loop and B is magnetic field

M = area x current

= 3.14 x .10² x i

Maximum torque = M B

1 x 10⁻³ = 3.14 x .10² x i x .65 x 10⁻⁴

i = 490 A

Current = 490 A.

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First box and third box !
4 0
3 years ago
The answer and how to do it?? Thanks
denis-greek [22]

Answer:

14 m/s²

Explanation:

Start with Newton's 2nd law: Fnet=ma, with F being force, m being mass, and a being acceleration. The applied forces on the left and right side of the block are equivalent, so they cancel out and are negligible. That way, you only have to worry about the y direction. Don't forget the force that gravity has the object. It appears to me that the object is falling, so there would be an additional force from going down from weight of the object. Weight is gravity (can be rounded to 10) x mass. Substitute 4N+weight in for Fnet and 1kg in for m.

(4N + 10 x 1kg)=(1kg)a

14/1=14, so the acceleration is 14 m/s²

4 0
3 years ago
Which planet is known as evening star​
Verdich [7]

Answer:

Mercury is a metal that is used in

Mercury can be seen as an evening "star" near the sun's setting point or as a morning "star" near the sun's rising point. The evening star was given the name Hermes, and the morning star was given the name Apollo, since the ancient Greeks thought they were two separate things. Mercury, the Roman god's messenger, is the planet's name.

8 0
3 years ago
What is the acceleration that earth experiences due to your gravitational pull?
djverab [1.8K]

Answer:

The acceleration that earth experiences due to  gravitational pull is = 9.81  \frac{m}{s^{2} }.

Explanation:

The acceleration that earth experiences due to  gravitational pull is called the acceleration due to gravity. its value is 9.81 and its unit is \frac{m}{s^{2} }.

When the object move upwards than in that case the earth gravitational force pulls down the body.

The formula of force due to gravity on the body is given as

F = mg

where g = acceleration due to gravity.

Due to this acceleration the body falls upon the surface of the earth.

4 0
3 years ago
A student pulls a sled of mass m = 82.0 kg with a force of F = 160N, and the force makes an angle of θ = 15 degrees with respect
Minchanka [31]

Answer:B

Explanation:

Given

mass of sledge=82 kg

Force on sledge=160 N

degree=15^{\circ}

force has two component sin and cos

Normal reaction =mg-F\sin \theta

N=82\times 9.8-160\sin 15

N=803.6-41.41=762.19 N

8 0
3 years ago
Read 2 more answers
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