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3 years ago

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A car is going through a dip in the road whose curvature approximates a circle of radius 200 m. At what velocity will the occupa

nts of the car appear to weigh 20% more than their normal weight (or their normal weight times 1.2)?

1 answer:

boyakko [2]3 years ago

4 0

Answer:

Explanation:

The solution along with the diagram can be found in the attachment. Feel free to raise doubts.

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A thin rod of length L and total charge Q has the nonuniform linear charge distribution λ(x)=λ0x/L, where x is measured from the

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Answer:

Explanation:

λ(x) = λo x/ L

(a) The total charge is Q.

$Q=\int_{0}^{L}dq$

$Q=\int_{0}^{L}\frac{\lambda _{0}x}{L}dx$

$Q=\frac{\lambda _{0}}{2L}\left ( x^{2} \right )_{0}^{L}$

$Q=\frac{\lambda _{0}}{2L}$

λo = 2Q/L

(b)

Let at a distance x from the origin the charge is dq.

so, dq = (2Q/L) x/ L dx

$dq=\frac{2Qx}{L^{2}}dx$

The potential due to this small charge at a distance d to the left of origin

$dV = \frac{KdQ}{d+x}$

$\int_{0}^{V}dV = \frac{2KQ}{L^{2}}\int_{0}^{L}\frac{xdx}{d+x}$

$V = \frac{2KQ}{L^{2}}\int_{0}^{L}\left ( 1- \frac{d}{d+x}\right )dx$

$V = \frac{2KQ}{L^{2}}\times \left ( x-dln(d+x) \right )\int_{0}^{L}$

$V = \frac{2KQ}{L^{2}}\times \left ( L-dln(d+L)-0+dlnd \right )$

$V = \frac{Q}{4\pi \epsilon _{0}L^{2}}\times \left ( L+d\times ln\left (\frac{d}{d+L} \right )\right )$

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True

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