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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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For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is
ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)

where A(t)=A₀e^{\frac{-bt}{2m}}

 A₀ is the amplitude at t=0 and

w' is the angular frequency of damped SHM, which is given by,

w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

Now coming to the problem,

Given: m=1.2 kg

           k=9.8 N/m

           b=210 g/s= 0.21 kg/s

           A₀=13 cm

a) A(t)=A₀/8

⇒A₀e^{\frac{-bt}{2m}} =A₀/8

⇒e^{\frac{bt}{2m}}=8

applying logarithm on both sides

⇒\frac{bt}{2m}=ln(8)

⇒t=\frac{2m*ln(8)}{b}

substituting the values

t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)

b) w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}

T'=\frac{2\pi}{w'}, where T' is time period of damped SHM

⇒T'=\frac{2\pi}{2.86}=2.2s

let n be number of oscillations made

then, nT'=t

⇒n=\frac{24}{2.2}=11(approx)

8 0
3 years ago
A single circular loop of wire of radius 0.75 m carries a constant current of 3.0 A. The loop may be rotated about an axis that
NISA [10]

Answer:

B = 0.8 T

Explanation:

It is given that,

Radius of circular loop, r = 0.75 m

Current in the loop, I = 3 A

The loop may be rotated about an axis that passes through the center and lies in the plane of the loop.

When the orientation of the normal to the loop with respect to the direction of the magnetic field is 25°, the torque on the coil is 1.8 Nm.

We need to find the magnitude of the uniform magnetic field exerting this torque on the loop. Torque acting on the loop is given by :

\tau=NIAB\sin\theta

B is magnetic field

B=\dfrac{\tau}{NIA\sin\theta}\\\\B=\dfrac{1.8}{1\times \pi \times (0.75)^2\times 3\times \sin(25)}\\\\B=0.8\ T

So, the magnitude of the uniform magnetic field exerting this torque on the loop is 0.8 T.

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Answer:

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The answer to the question is A 
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Why are compounds containing carbon called organic chemicals
sergey [27]

Answer:

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Explanation:

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