Let us situate this on the x axis, and let our uniform line of charge be positioned on the interval <span>(−L,0]</span> for some large number L. The voltage V as a function of x on the interval <span>(0,∞)</span> is given by integrating the contributions from each bit of charge. Let the charge density be λ. Thus, for an infinitesimal length element <span>d<span>x′</span></span>, we have <span>λ=<span><span>dq</span><span>d<span>x′</span></span></span></span>.<span>V(x)=<span>1/<span>4π<span>ϵ0</span></span></span><span>∫line</span><span><span>dq/</span>r</span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>∫<span>−L</span>0</span><span><span>d<span>x/</span></span><span>x−<span>x′</span></span></span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>(ln|x+L|−ln|x|)</span></span>
Answer: Power is 200 W
Explanation: Power P = work done / time used.
P = W/t = mgh/t = 154 kg · 9.81 m/s²· 4 m / 30 s = 201 W
Answer:
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Answer:
(a) 8.362 rad/sec
(b) 6.815 m/sec
(c) 9.446
(d) 396.22 revolution
Explanation:
We have given that diameter d = 1.63 m
So radius
Angular speed N = 79.9 rev/min
(a) We know that angular speed in radian per sec
(b) We know that linear speed is given by
(c) We have given final angular velocity
And
Time t = 63 sec
Angular acceleration is given by
(d) Change in angle is given by