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AlekseyPX
3 years ago
6

Starting from rest, a wheel with constant angular acceleration turns through an angle of 25 rad in a time t. Through what angle

will it have turned after time 2t?
Physics
1 answer:
WITCHER [35]3 years ago
8 0

Answer:

After time 2t, the wheel will turn through an angle of 100 rad.

Explanation:

Given;

initial velocity of the wheel, \omega_i = 0

time of motion, t = t

angular distance, θ = 25 rad

The constant angular acceleration is calculated as;

\theta = ut + \frac{1}{2}at^2\\\\\theta = 0 + \frac{1}{2}at^2\\\\2\theta = at^2\\\\a = \frac{2\theta }{t^2} \\\\a = \frac{2\ \times \ 25 }{t^2}\\\\a = \frac{50}{t^2} \\\\when \ t = 2t\\\\\theta = \frac{1}{2}at^2\\\\\theta = \frac{1}{2}(\frac{50}{t^2})(2t)^2\\\\\theta =\frac{1}{2}(\frac{50}{t^2})(4t^2)\\\\\theta = 100 \ rad

Therefore, after time 2t, the wheel will turn through an angle of 100 rad.

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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.

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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}.

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  • A = \rm 71\; cm^{2} = 71\times \left(10^{-2}\right)^2 \; m^{2} is the area of the coil.
  • N = 1 is the number of loops in the coil.
  • \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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