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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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Read 2 more answers

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$\Phi_B(t)=NABcos(\omega t)\\\\\omega=\frac{2\pi}{T}=1.25\frac{rad}{s}\\\\A=l^2=(0.02m)^2=4*10^{-4}m^2\\\\B=0.1T\\\\\Phi_B(t)=1.2*10^{-3}cos(1.25 t) W$

where we have used the values given by the information of the problem for N B and A.

the emf is given by:

$emf=-\frac{d\Phi_B}{dt}=-NBA\omega sin(\omega t)\\\\emf(t=12.5s)=-(30)(0.1T)(4*10^{-4})(1.25\frac{rad}{s})sin(1.25*12.5)=1.49*10^{-4}V=0.49mV$

hope this helps!!

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