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Sphinxa [80]
3 years ago
11

A ball dropped from rest falls freely until it hits the ground with a speed pf 20 m/s. The time during with the ball is in free

fall is approximately
Physics
1 answer:
saul85 [17]3 years ago
5 0

Answer:

\large \boxed{\text{2.0 s}}

Explanation:

\begin{array}{rcl}v & = & gt\\\text{20 m$\cdot$s}^{-1} & = & \text{9.807 m$\cdot$ s}^{-2} \times t\\t & = & \dfrac{20}{9.807 \text{ s}^{-1}}\\\\& = & \textbf{2.0 s}\\\end{array}\\\text{The ball is in free fall for $\large \boxed{\textbf{2.0 s}}$}

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A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the
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The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

E=\dfrac{Q}{4\pi\epsilon_0 r^2}

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The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

Learn more about Gaussian sphere here:

brainly.com/question/2004529

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1 year ago
TESE
Dmitry [639]

equal and opposite reaction.

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3 years ago
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