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vazorg [7]
3 years ago
9

Does the voltage of a battery affect the strength of an electromagnet?

Physics
1 answer:
swat323 years ago
7 0

I'm trying to make an electromagnet that's strength is constantly getting incremented by small amounts every second. I need to know, which would have a greater effect on the electromagnet's strength, amps or volts? (I know increasing the turns and/or density of the magnet wire will increase the strength, but I am looking for answers other than that particular one.)

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On Monday, Jacques announces to his parents that he wants to be called "Jack." On Wednesday, he says he wants to drop out of sch
Alex777 [14]
Jaques seems to be in the Rebelious stage
5 0
3 years ago
In some cases fixture wires may be used for
zalisa [80]

You can use fixture wires: For installation in luminaires where they are enclosed and protected and not subject to bending and twisting and also can be used to connect luminaires to their branch circuit conductors.

<h3>What are some uses of fixture wires?</h3>

Fixture wires are flexible conductors that are used for wiring fixtures and control circuits. There are some special uses and requirements for fixture wires and no fixture can be smaller than 18 AWG

In modern fixtures, neutral wire is white and the hot wire is red or black. In some types of fixtures, both wires will be of the same color.

To know more about fixture wires, refer

brainly.com/question/26098282

#SPJ4

3 0
1 year ago
A light ray incident on a block of glass makes an incident angle of 50.0° with the normal to the surface. The refracted ray in t
Vladimir [108]

Answer:

The index of refraction of the glass is 1.3

Explanation:

Given data:

i = incident angle = 50°

r = refracted angle = 36.1°

The index of refraction according Snell´s law is:

n=\frac{1*sini}{sinr} =\frac{1*sin50}{sin36.1} =1.3

4 0
3 years ago
A horizontal force of 140 N is needed to pull a 60kg box across the horizontal floor at constant speed . Find the coefficient of
____ [38]

Explanation:

formula for force is:

force=mass × acceleration

but in case of friction

force =coefficient of friction × Normal Reaction

F. = u × R

U = F/R

but when placed horizontally

R= M×G

M=mass=60kg

G=Gravity(10m/s or 9.8m/s)

F=140N

U=140/60×10

U=140/600

U=0.2333333333

approximately to 3 significant figures

U=0.233

if i am correct rate it 5 star

4 0
1 year ago
Suppose we take a 1 m long uniform bar and support it at the 33 cm mark. Hanging a 0.15 kg mass on the short end of the beam res
MakcuM [25]

Answer:

The mass of the beam is 0.074 kg

Explanation:

Given;

length of the uniform bar, = 1m = 100 cm

Set up this system with the given mass and support;

  0-----------------33cm-----------------------------------100cm

  ↓                     Δ                                             ↓      

0.15kg                                                              m

Where;

m is mass of the uniform bar

Apply the principle of moment to determine the value of "m"

sum of anticlockwise moment = sum of clockwise moment

 0.15kg(33 - 0) = m(100 - 33)

0.15(33) = m(67)

m = \frac{0.15kg(33 \ cm)}{67 \ cm}\\\\m = 0.074 \ kg

Therefore, the mass of the beam is 0.074 kg

8 0
3 years ago
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