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2 years ago

A student drops a rock from a bridge to the water 12m below. How many seconds does it take the rock to hit the water?

1 answer:

7 0

R = $\frac{g t^{2} }{2}$
r - displacement or height
t - time taken
g = 10 m/s-2
t = square root from 2r/g = $\sqrt{ \frac{2 x 12}{10} }$ = 1.5 seconds

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An uncharged, nonconducting, hollow sphere of radius 10.0cm surrounds a 10.0-μC charge located at the origin of a Cartesian coor

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The electric flux through the hole is $56.45\ webber$ .

Electric flux is the number of electric field lines cutting through the surface and is measured as surface intregal of electric field over that surface
Mathematically it is given by $\phi_E=E.A \ Nm^2/C$ where E is the electric field and A is the area.
Gauss's law states that electric flux through closed surface is equal to the 1 / ε₀ times the charge enclosed by that surface which is given by Ф = q / ε₀ where q is the central charge and ε₀ is the permittivity of the medium.

It is given , hollow sphere of radius 10.0cm surrounds a 10.0-μC charge.

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$= 4\times 3.14\times (10 \times 10^{-2})^2 \\\\= 12.56\times 10^{-2} m^2$

Area of the hole ( both side ) $= 2\times \pi r^2$

$= 2\times 3.14 \times (10^-^3)^2\\= 6.28 \times 10^-^6 m^2$

According to Gauss's theorem, the flow from a particular charge in the center is given by

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This flux flows through the surface of the sphere, so the flux per unit area which is given by

$= \frac{ 1.13\times 10^6 }{ 12.56\times 10^-^2} \\\\= 8.99 \times 10^6 \ weber / m^2$

Flux through area of hole is given by :

$= 8.99\times10^6 \times6.28 \times 10^-^6\\ = 56.45 \ weber$

Learn about more electric flux here :

brainly.com/question/26289097

#SPJ4

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