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ozzi
4 years ago
15

The electric current running through the wire coil in an electric motor exerts force directly onto A) the battery. B) an aluminu

m axle. C) a powerful magnet. D) a rubber insulator.
Physics
2 answers:
antiseptic1488 [7]4 years ago
8 0
<span>C)<span>a powerful magnet.

</span></span>
Jobisdone [24]4 years ago
3 0

Answer:

C) a powerful magnet

Explanation:

The electric current running through the wire coil in an electric motor exerts force directly onto A) the battery. B) an aluminum axle. C) a powerful magnet. D) a rubber insulator.

when current is pass to the coil, an electromagnetic force field is produced. The electromagnetic force field produced then have an impact on the magnet which drives the shaft to move

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How much force is required to accelerate a 12 kg mass at 5 m/s 2
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Answer:

60 N

Explanation:

This is just Newton's Second Law

F = m*a

F = ?

m = 12 kg

a = 5 m/^2

F = 5*12 = 60 Newtons

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Lab: Motion with Constant Acceleration Assignment: Lab Report
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3 years ago
Of the following gases, ________ will have the greatest rate of effusion at a given temperature.
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3 years ago
Read 2 more answers
Find the area under the standard normal curve between z=0.19 and z=2.18. round your answer to four decimal places, if necessary.
LiRa [457]

The area under the standard normal curve between z=0.19 and z=2.18 is,

P(0.19<z<2.18)= 0.4101

<h3>How can we calculate the area under the curve?</h3>

To calculate the The area under the standard normal curve  between z=0.19 and z=2.18, we are using two things,

<u>Step 1</u>: The formula,

P(0.19<z<2.18)= P(Z<2.18)- P(Z<0.19)

<u>Step 2</u>: The statistical values of the area under the curve we get from the picture.

Now we put the known values from the picture in the above formula, we get,

P(0.19<z<2.18)= P(Z<2.18)- P(Z<0.19)

Or, P(0.19<z<2.18)=  0.9854-0.5753

Or, P(0.19<z<2.18)=  0.4101

From the above calculation we can easily conclude that,

The area under the standard normal curve between z=0.19 and z=2.18 is,

P(0.19<z<2.18)= 0.4101

Learn more about the standard normal curve:

brainly.com/question/17088387

#SPJ4

5 0
2 years ago
An automobile accelerates from zero to 30 m/s in 6.0 s. The wheels have a diameter of 0.40 m. What is the average angular accele
leva [86]

To solve this problem we will use the concepts related to angular motion equations. Therefore we will have that the angular acceleration will be equivalent to the change in the angular velocity per unit of time.

Later we will use the relationship between linear velocity, radius and angular velocity to find said angular velocity and use it in the mathematical expression of angular acceleration.

The average angular acceleration

\alpha = \frac{\omega_f - \omega_0}{t}

Here

\alpha = Angular acceleration

\omega_{f,i} = Initial and final angular velocity

There is not initial angular velocity,then

\alpha = \frac{\omega_f}{t}

We know that the relation between the tangential velocity with the angular velocity is given by,

v = r\omega

Here,

r = Radius

\omega = Angular velocity,

Rearranging to find the angular velocity

\omega = \frac{v}{r}}

\omega = \frac{30}{0.20} \rightarrow Remember that the radius is half te diameter.

Now replacing this expression at the first equation we have,

\alpha = \frac{30}{0.20*6}

\alpha = 25 rad /s^2

Therefore teh average angular acceleration of each wheel is 25rad/s^2

3 0
3 years ago
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