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Pepsi [2]
2 years ago
14

What question is maya most likely trying to answer? how does the type of material used in the core affect the strength of an ele

ctromagnet? how does the number of loops of wire affect the strength of an electromagnet? how does the speed at which an iron bar is removed from wire loops affect the strength of an electromagnet? how does the thickness of the wire affect the strength of an electromagnet?
Physics
1 answer:
SashulF [63]2 years ago
4 0

The question is maya most likely trying to answer is, how does the thickness of the wire affect the strength of an electromagnet?

<h3>What is effect of thickness of wire on strength of electromagnet?</h3>

From the experimental set-up by maya, we can determine the effect of thickness of wire on strength of electromagnet.

R ∝ 1/A

where;

  • R is resistance of the wires
  • A is area of the wires (from thickness or radius of the wire)

As the thickness of the wire increases, the area of the wire increases and the resistance of the wire will decrease. As the resistance of the wire decreases, the current flowing in the wire increases as well.

Thus, the increase in the thickness of a wires increases, the strength of an electromagnet.

Learn more about strength of an electromagnet here: brainly.com/question/2331156

#SPJ4

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Explain how the fountain pen is filled up with ink​
Leno4ka [110]

Answer: A piston-filling fountain pen has a piston — just like in a car — inside the barrel. This piston goes down to expel air or ink and then back up, pulling ink into the barrel. The typical process is very simple, assuming the pen is clean and dry: Push the piston down, expelling any air in the barrel

8 0
3 years ago
A guy wire 1034 feet long is attached to the top of a tower. When pulled taut, it touches level ground 699 feet from the base of
kolezko [41]

Answer:

80.386 degrees

Explanation:

We use the cosine equation here (which is the adjacent side of the unknown angle divided by the hypotenuse

The adjacent side = 699ft

The hypotenuse = 1034ft

using cos∅ = Adjacent/hypotenuse

where ∅ is the unknown angle

cos ∅ = 699/1034 = 0.167

∅ = arccos 0.167 = 80.368°

As easy as one can imagine

8 0
3 years ago
Which statement about an atom is correct?
Alexus [3.1K]

Answer:

\boxed{ \sf{The \: electron \: has \: a \: negative \: charge \: and \: is \: found \: outside \: of \: the \: nucleus}}

Option A is correct

Explanation:

Let's check the options:

A: The electron has a negative charge and is found outside of the nucleus.

\mapsto{} Yeah! It's TRUE . An electron is a <u>negatively</u><u> </u><u>charged</u><u> </u><u>particle</u><u> </u><u>and</u><u> </u><u>is</u><u> </u><u>located</u><u> </u><u>outside</u><u> </u><u>the</u><u> </u><u>nucleus</u><u> </u><u>of</u><u> </u><u>an</u><u> </u><u>atom.</u>

B : The neutron has a negative charge and is found in the nucleus.

\mapsto{}No! It's FALSE . A neutron carries <u>no </u><u>charge</u>. i.e it is a neutral particle and found inside the nucleus.

C : The proton has no charge and is found in the nucleus.

\mapsto{}No! It's FALSE.A proton is a <u>positively</u><u> </u><u>charged</u><u> </u><u>particle</u> present inside nucleus of an atom.

D : The neutron has no charge and is found outside of the nucleus.

\mapsto{} I agree that the neutron has no charge. But it is found <u>inside</u> the nucleus not outside . So, this statement is FALSE .

Hence, we found our answer! :D

A. The electron has a negative charge and is found outside of the nucleus is the correct statement about an atom.

Hope I helped!

Best regards! :D

~\text{TheAnimeGirl}

7 0
3 years ago
Photons of wavelength 65.0 pm are Compton-scattered from a free electron which picks up a kinetic energy of 0.84 keV from the co
ElenaW [278]

Answer:

λ  = 65.6 pm

Explanation:

Given that

λo = 65 pm

The initial energy of the electron

E_o=\dfrac{hC}{\lambda_0 }

Now by putting the values

E_o=\dfrac{hC}{\lambda_0 }

E_o=\dfrac{6.67\times 10^{-34}\times 3\times 10^8}{65\times 10^{-12}}

E_o=3.05\times 10^{-15}\ J

E_o=\dfrac{3.05\times 10^{-15}}{1.6\times 10^{-19}\times 10^3}\ KeV

Eo=19.06 KeV

Given that kinetic energy KE= 0.84 KeV

Therefore the final energy

E= Eo - KE

E = 19.06 - 0.84 KeV

E= 18.22 KeV

The wavelength  λ can be find as

E=\dfrac{hC}{\lambda}

\lambda=\dfrac{hC}{E}

\lambda=\dfrac{6.67\times 10^{-34}\times 3\times 10^8}{19.06 \times 10^3\times 1.6\times 10^{-19}}

λ = 6.56 x 10⁻¹¹ m

λ  = 65.6 pm

3 0
3 years ago
How do salt and water form a solution?
iogann1982 [59]
It would be D. The solute, salt, must dissolve IN the solvent, water (:
8 0
4 years ago
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