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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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Explanation:

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How much time does it take for a car to accelerate from 3.44 m/s to 20.9 m/s if the average acceleration is 6.00 m/s^3
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A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m
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Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, I_1=900\ kg.m^2

Initial angular velocity of the platform, \omega=0.95\ rad/s

Part A,

Let \omega_2 is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

I_1\omega_1=I_2\omega_2

Here, I_2=I_1+mr^2

I_1\omega_1=(I_1+mr^2)\omega_2

900\times 0.95=(900+70\times (2.9)^2)\omega_2

Solving the above equation, we get the value as :

\omega_2=0.574\ rad/s

Part B,

The initial rotational kinetic energy is given by :

k_i=\dfrac{1}{2}I_1\omega_1^2

k_i=\dfrac{1}{2}\times 900\times (0.95)^2

k_i=406.12\ rad/s

The final rotational kinetic energy is given by :

k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2

k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2

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3 years ago
A truck accelerates From being stopped to 18 m/s in 5.5 seconds. A bus accelerates from being stopped to 24 m/s in 6.0 seconds.
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Answer:

The bus

Explanation:

a = (v-u)/t  

Where

a = acceleration

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u = initial velocity

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For bus to get its acceleration,

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