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Bezzdna [24]
4 years ago
14

A wheel has a constant angular acceleration of 4.5 rad/s2. during a certain 5.0 s interval, it turns through an angle of 128 rad

. assuming that the wheel started from rest, how long has it been in motion at the start of this 5.0 s interval?
Physics
1 answer:
dalvyx [7]4 years ago
3 0
The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2 
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration 
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1 
w1 = wo + ãt 
w1 - final angular velocity 
wo - initial angular velocity 
18 = 0 + 4.5t t = 4 s
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The solution to the questions are given as

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  • \varepsilon &=(0.12v)e^{0.057t}
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<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

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\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}

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(b) Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$

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