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Nitella [24]
3 years ago
11

The electric current running through the wire coil in an electric motor exerts force directly onto

Physics
2 answers:
Sveta_85 [38]3 years ago
7 0
C) A powerful magnet
Mashcka [7]3 years ago
6 0

The correct choice is

C) a powerful magnet.

we know that a current carrying wire creates magnetic field around it. inside this magnetic field, when a magnet comes, the magnet experience magnetic force. This same thing happen in motor which is known as motor effect.

the electric current running through coil creates magnetic field which exerts the force on the permanent magnet.


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There are 6 types of quarks.

Answer: C. 6
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3 years ago
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Fraternity hazing is acceptable because it is an initational rite to the brotherhood
Fantom [35]

Answer:

that is the right answer

Explanation:

8 0
3 years ago
the magnitude of the magnetic field at point p for a certain electromagnetic wave is 2.21. What is the magnitude of the elctic f
vesna_86 [32]

Answer:

6.63\times 10^8\ N/C

Explanation:

Given that,

The magnitude of magnetic field, B = 2.21

We need to find the magnitude of the electric field. Let it is E. So,

\dfrac{E}{B}=c\\\\E=Bc

Put all the values,

E=2.21\times 3\times 10^8\\\\=6.63\times 10^8\ N/C

So, the magnitude of the electric field is equal to 6.63\times 10^8\ N/C.

7 0
3 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
3 years ago
Please Help!
irga5000 [103]

The maximum force that the tires can exert on the road before slipping is 16200 N.

From the information in the question;

The coefficient of static friction =  0.9

The mass of the car = 1800 kg

Using the formula;

μ = F/R

μ  = coefficient of static friction

F = force on the tires

R = the reaction force

But recall that the reaction is equal in magnitude to the weight of the car.

W=R

Hence; R = 1800 kg × 10 ms-2 = 18000 N

Making F the subject of the formula;

F = μR

Substituting values;

F =  18000 N × 0.9

F = 16200 N

Hence, the maximum force that the tires can exert on the road before slipping is 16200 N.

Learn more: brainly.com/question/18754989

6 0
3 years ago
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