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2 years ago

What is the IMA of the following pulley system?

Physics

2 answers:

Dmitry [639]2 years ago

6 0

Answer : The IMA of the given pulley system is $\dfrac{F_r}{F_e}$ .

Explanation :

The full form of IMA is an ideal mechanical advantage. For the pulley system, it is defined as the ratio of the output force (resistance force) and the input force (effort force).

In this case, the input force is $F_e$

and the output force is $F_r$

Mathematically, the IMA for the pulley system it can be written as:

$IMA=\dfrac{F_r}{F_e}$

Hence, this is the required solution.

Furkat [3]2 years ago

5 0

Answer: $IMA= \frac {F_r}{F_e}$

Explanation:

IMA stands for Ideal Mechanical advantage.

The IMA of pulley system can be defined as the ratio of output force to input force.

From the given pulley system,

The input force = $F_e$

The output force = $F_r$

Hence, IMA of the given pulley system,

$IMA= \frac {F_r}{F_e}$

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AysviL [449]

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zzz [600]

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3 years ago

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kkurt [141]

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

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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.
$N = 1$ 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, $\displaystyle \theta = 90^{\circ}$ .