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FinnZ [79.3K]
3 years ago
15

A square metal block with mass (m) is suspended by five wires with the same length. One of these wires is made of iron at the ce

nter of the block and has a radius (1.2 mm) while the other four wires have the same cross-sectional area and they are made of copper at the block corners. If the tension in the iron wire equals (0.3 mg), find the radius of a copper wire. (Take Yiron = 2×1011, Ycopper =1. 2×1011 N/m2 )
Physics
1 answer:
blsea [12.9K]3 years ago
5 0

Answer:

1.18 mm

Explanation:

The mass of the block is m, so the weight of the block is mg.  If the tension in the iron wire is 0.3 mg, then the total tension in the 4 copper wires is 0.7 mg.  So the tension in each copper wire is 0.175 mg.

For each wire:

ΔL/L = F / (EA)

where ΔL is the change in length,

L is the initial length,

F is the force,

E is Young's modulus (modulus of elasticity),

and A is the cross sectional area.

The five wires have the same initial length, and stretch by the same amount.

ΔL/L = ΔL/L

F / (EA) = F / (EA)

(0.3 mg) / (2×10¹¹ Pa × π(1.2 mm)²) = (0.175 mg) / (1.2×10¹¹ Pa × πr²)

0.3 / (2 (1.2 mm)²) = 0.175 / (1.2 r²)

0.36 r² = 0.504 mm²

r = 1.18 mm

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Aloop of wire of area 71 cm^2 is placed with its plane parallel to a 16 mt magnetic field. the loop is then rotated so that its
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Answer:

Approximately 1.62 × 10⁻⁴ V.

Explanation:

The average EMF in the coil is equal to

\displaystyle \frac{\text{Final Magnetic Flux} - \text{Initial Magnetic Flux}}{2},

Why does this formula work?

By Faraday's Law of Induction, the EMF \epsilon induced in a coil (one loop) is equal to the rate of change in the magnetic flux \Phi through the coil.

\displaystyle \epsilon(t) = \frac{d}{dt}(\Phi(t)).

Finding the average EMF in the coil is similar to finding the average velocity.

\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt.

However, by the Fundamental Theorem of Calculus, integration reverts the action of differentiation. That is:

\displaystyle \int_0^{t} \epsilon(t)\cdot dt = \int_0^{t} \frac{d}{dt}\Phi(t)\cdot dt = \Phi(t) - \Phi(0).

Hence the equation

\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt = \frac{\Phi(t)- \Phi(0)}{t}.

Note that information about the constant term in the original function will be lost. However, since this integral is a definite one, the constant term in \Phi(t) won't matter.

Apply this formula to this question. Note that \Phi, the magnetic flux through the coil, can be calculated with the equation

\Phi = B \cdot A \cdot N \; \sin{\theta}.

For this question,

  • B = \rm 16\; mT = 16\times 10^{-3}\; T is the strength of the magnetic field.
  • A = \rm 71\; cm^{2} = 71\times \left(10^{-2}\right)^2 \; m^{2} is the area of the coil.
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  • \theta is the angle between the field lines and the coil.
  • At \rm 0\;s, the field lines are parallel to the coil, \theta = 0^{\circ}.
  • At \rm 0.7\; s, the field lines are perpendicular to the coil, \displaystyle \theta = 90^{\circ}.

Initial flux: \Phi(0)= 0.

Final flux: \Phi(0.7) = \rm 1.1136\times 10^{-4}\; Wb.

Average EMF, which is the same as the average rate of change in flux:

\displaystyle \frac{\Phi(0.7) - \Phi(0)}{0.7} \approx\rm 1.62\times 10^{-4}\; V.

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A chamber fitted with a piston contains 1.90 mol of an ideal gas. Part A The piston is slowly moved to decrease the chamber volu
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Answer:

The work done is 5136.88 J.

Explanation:

Given that,

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Put the value into the formula

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W=1.90\times8.314\times296\ ln(\dfrac{1}{3})

W=−5136.88\ J

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Hence, The work done is 5136.88 J.

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