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3 years ago

How does wire thickness affect the power of the electromagnet?

1 answer:

6 0

It does not depend on the thickness of the wire. Well, not directly.

As steveastrouk wrote, the magnetic field depends on the current through the wire and how often the wire goes around the core - the number of loops.

If you use a thinner wire, you can wrap more turns into a given space. Good for the magnetic field! But...

A thinner wire means more resistance. More turns mean a longer wire and that means even more resistance. More resistance means that a given voltage can drive less current through the wire. Less current means a lower magnetic field.

If and when the two effects cancel each other, well that's what math is for.

(Of course you could increase the voltage to keep the current in the thinner wire the same as in the thicker wire, but that would increase the thermal losses in the wire as well.)

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In an experiment the mass of a calorimeter is 36.35 g . Express in micrometer ,millimetre and kg.

Andru [333]

Answer:

1. 36.35 g = 36.35E15 micrometer.

II. 36.35 g = 363.5 millimetre.

III. 36.35 g = 0.03635 kilogram.

Explanation:

Given the following data;

Mass of calorimeter = 36.35 grams

To convert the mass in grams (g) to;

I. Micrometer

Conversion:

1 g = 1 exp 15 um

36.35 g = X um

Cross-multiplying, we have;

X = 36.35 * 1 exp 15 = 36.35 exp 15 um

36.35 g = 36.35E15 micrometer

II. Millimetre

Conversion:

1 g = 1 milliliter

36.35 g = X milliliter

Cross-multiplying, we have;

X = 36.35 * 1 = 36.35 milliliter

Next, we would convert milliliter to millimetre;

1 milliliter = 10 millimetre

36.35 milliliter = X millimetre

Cross-multiplying, we have;

X = 36.35 * 10 = 363.5 millimetre

36.35 g = 363.5 millimetre

III. Kilogram

Conversion:

1000 grams = 1 kilogram

36.35 g = X kilogram

Cross-multiplying, we have;

X * 1000 = 36.35 * 1

Dividing both sides by 1000, we have;

X = 36.35/1000 = 0.03635 kilogram

36.35 g = 0.03635 kilogram

Note:

g is the symbol for grams.
Exp (E) means exponential = 10
um is the symbol for micrometer.

4 0

3 years ago

What is the weight of an astronaut who has a mass of 90 kg on the moon? (Note: acceleration due to gravity of the moon is 1.62 N

motikmotik

Answer:

W = 145.8 [N]

Explanation:

To solve this problem we must remember that weight is defined as the product of mass by gravity, in this case lunar gravity.

W = m*g

where:

m = mass = 90 [kg]

g = gravity acceleration = 1.62 [kg/m²]

W = 90*1.62

W = 145.8 [N]

8 0

3 years ago

According to research, college graduates are more likely to engage in all of the following EXCEPT:

Ksivusya [100]

Strange question. I’d say B)

7 0

3 years ago

Read 2 more answers

as your roller coaster climbs to the top of the steepest hill on its track when does the first car have the greatest potential e

pochemuha

...the potential energy that you build while going up the hill on the roller coaster could be let go as kinetic energy -- the energy of motion that takes you down the hill of the roller coaster.

4 0

3 years ago

A student places an object with a mass of m on a disk at a position r from the center of the disk. The student starts rotating t

koban [17]

Answer:

The coefficient of static friction between the object and the disk is 0.087.

Explanation:

According to the statement, the object on the disk experiments a centrifugal force due to static friction. From 2nd Newton's Law, we can represent the object by the following formula:

$\Sigma F_{r} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}$ (1)

$\Sigma F_{y} = N - m\cdot g = 0$ (2)

Where:

$N$ - Normal force from the ground on the object, measured in newtons.

$m$ - Mass of the object, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v$ - Linear speed of rotation of the disk, measured in meters per second.

$R$ - Distance of the object from the center of the disk, measured in meters.

By applying (2) on (1), we obtain the following formula:

$\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}$

$\mu_{s} = \frac{v^{2}}{g\cdot R}$

If we know that $v = 0.8\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $R = 0.75\,m$ , then the coefficient of static friction between the object and the disk is:

$\mu_{s} = \frac{\left(0.8\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.75\,m)}$

$\mu_{s} = 0.087$

The coefficient of static friction between the object and the disk is 0.087.

5 0

3 years ago