solution:
E\delta =\frac{R}{\epsilon0}(1-\frac{A}{\sqrt{4R^{2}}-ac}
=\frac{R}{\epsilon0}(1-\frac{1}{\sqrt{4r^{2}/^{_a{2}}+1}})
=\frac{R}{\epsilon0}(1-\frac{1}{\sqrt{4x^2+1}})
x=\frac{r}{a}
infinite case,
Ei=\frac{r}{\epsilon0}
\therefore e\delta =ei(1-\frac{1}{\sqrt{4x^{2}+1}})
we have to find x when,
ei-e\delta =1% ,y=ei=1/100 ei
or,ei-ei+\frac{ei}{\sqrt{4x^2+1}} = 1/100ei
\frac{1}{\sqrt{4x^2+1}}=\frac{1}{100}
4x^2+1 =10^4
x=\frac{\sqrt{\frac{10^4-1}{4}}}=49.99\approx 50
\therefore \frac{r}{a}\approx 50
Answer with Explanation:
We are given that
Mass of rock=m
Maximum height=h
a.At maximum height, velocity,v=0
We know that
Height,h=h/4
Again,
Where
b.When height,h=3h/4
Answer:
Diameter of a sphere: d = 2 * r , Surface area of a sphere: A = 4 * π * r² , Volume of a sphere: V = 4/3 * π * r³ , Surface to volume ratio of a sphere: A / V = 3 / r . To Measure Diameter of a Small Spherical/ Cylindrical Body Using Vernier's Callipers.
Explanation:
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