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3 years ago

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A solenoid used to produce magnetic fields for research purposes is 2.1 m long, with an inner radius of 28 cm and 1000 turns of

wire. When running, the solenoid produced a field of 1.3 T in the center. Given this, how large a current does it carry?

1 answer:

Veseljchak [2.6K]3 years ago

7 0

Answer:

I = 2172.46 A

Explanation:

Given that,

The length of a solenoid, l = 2.1 m

The inner radius of the solenoid, r = 28 cm = 0.28 m

The number of turns in the wire, N = 1000

The magnetic field in the solenoid, B = 1.3 T

We need to find the current carried by it. We know that, the magnetic field in a solenoid is given by :

$B=\mu_o nI\\\\or\\\\B=\mu_o \dfrac{N}{L}I\\\\I=\dfrac{BL}{\mu_o N}$

Put all the values,

$I=\dfrac{1.3\times 2.1}{4\pi \times 10^{-7}\times 1000}\\\\I=2172.46\ A$

So, it carry current of 2172.46 A.

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If the coefficient of kinetic friction between a 22 kg kg crate and the floor is 0.27, what horizontal force is required to move

alina1380 [7]

Answer:

58.27 N

Explanation:

the data we have is:

mass: $m=22kg$

coefficient of friction: $\mu =0.27$

and we also know the acceleration of gravity is $g=9.81m/s^2$

We need to do an analysis of horizontal and vertical forces acting on the object:

-------

Vertically the forces acting on the object:

Normal force $N$ (acting up from the object)

weight: $w=mg$ (acting down from)

so the sum of forces in the vertical axis "y" are:

$F_{y}=N-w\\F_{y}=N-mg$

from Newton's second Law we know that $F=ma$ , so:

$ma_{y}=N-mg$

and since the object is not accelerating in the vertical direction (the movement is only horizontal) $a_{y}=0$ , and:

$0=N-mg\\N=mg$

-----------

now let's analyze the horizontal forces

frictional force: $f= \mu N$ and since $N=mg$ --> $f=\mu mg$

force to move the object: $F$

and the two forces just mentioned must be opposite, thus the sum of forces in the "x" axis is:

$F=ma_{x}=F-f\\ma_{x}=F-\mu mg$

and we are told that the crate moves at a steady speed, thus there is no acceleration: $a_{x}=0$

and we get:

$0=F-\mu mg\\F=\mu mg$

substituting known values:

$F=(0.27)(22kg)(9.81m/s^2)\\F=58.27N$

3 0

3 years ago

A weightlifter lifts a 1250-N barbell 2 m in 3 s.How much power was used to lift the barbell?

Vlad [161]

The power is 833.3 W

Explanation:

First of all, we need to calculate the work done in lifting the barbell, which is equal to the change in gravitational potential energy of the barbell:

$W=(mg)h$

where

mg = 1250 N is the weight of the barbell

h = 2 m is the change in height

Substituting,

$W=(1250)(2)=2500 J$

Now we can calculate the power, which is equal to the work done per unit time:

$P=\frac{W}{t}$

where

W = 2500 J is the work done

t = 3 s is the time taken

Substituting,

$P=\frac{2500}{3}=833.3 W$

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

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3 years ago

a ball of mass 0.5 kg moving at 10 m/s collides with another ball of equal mass at rest. if the two balls move off together afte

nexus9112 [7]

Answer:

5 m/s

Explanation:

Here we can see there is no external force acted on a two masses when we consider the motion. If there is no external forces then momentum is conserved.

Initial momentum = Final momentum

0.5 × 10 = 1 × V

V = 5 m/s

7 0

3 years ago

What is the mass of an object with a density of 18 g/cm3 and a volume of 25 cm3

DiKsa [7]

Answer: 0.72 grams

Explanation: Mass can be extracted from the formula of density. D=M/V where D is density and V is volume. Therefore:

18 g/cm^3 = M(25 cm^3) --> Divide by 18g/cm^3 by 25 cm^3 to isolate mass. --> <u>0.72 =M </u> --> Now, to find out which unit you need to use for mass, just look at the density. You can see it is in g/cm^3, and cm^3 was already used for the volume. Thus, gram units are left, so that will be the unit needed, making the final answer 0.72 grams. Hope this helps :)

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2 years ago

n a car lift used in a service station, compressed air exerts a force on a small piston that has a circular cross section of rad

Oksi-84 [34.3K]

Answer: 1477.78 N

Explanation:

Let's assume that the cross sectional area of the smaller piston be A1

let's also assume the cross sectional area of the larger piston be A2

We assume the force applied to the smaller piston be F1

We also assume the force applied to the larger piston be F2

we then use the formula

F1/A1 = F2/A2

From our question,

The radius of the smaller piston is 5 cm = 0.05 m

The radius of the larger piston is 15 cm = 0.15 m

The force of the larger piston is 13300 N

The force of the smaller piston is unknown = F

A1 = πr² = 3.142 * 0.05² = 0.007855 m²

A2 = πr² = 3.142 * 0.15² = 0.070695 m²

F1/0.007855 = 13300/0.070695

F1 = (13300 * 0.007855) / 0.070695

F1 = 104.4715 / 0.070695

F1 = 1477.78 N

Thus, the force the compressed air must exert is 1477.78 N

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3 years ago