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3 years ago

1. An electron travels 4.82 meters in 0.00360 seconds. What is its average speed?

Physics

1 answer:

vredina [299]3 years ago

6 0

Answer:

speed =distance /time

speed =4.82/0.00360

speed =1338.8m/s

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2 years ago

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 57.3 c

Mashcka [7]

Answer with Explanation:

We are given that

Diameter of pipe, $d_1=0.573 m$

$v_1=13.5 m/s$

$v_2=5.83 m/s$

Volume flow rate of the petroleum along the pipe= $Q_{refinery}=A_1v_1=v_(\frac{\pi d^2_1}{4})$

$Q_{refinery}=13.5\times (\pi\times \frac{0.573)^2}{4})=3.48 m^3/s$

By equation of continuity

$A_1v_1=A_2v_2$

$\frac{\pi d^2_1}{4}v_1=\frac{\pi d^2_2}{4}v_2$

$d^2_2=\frac{v_1}{v_2}d^2_1$

$d_2=\sqrt{\frac{v_1}{v_2}}d_1$

$d_2=0.573\sqrt{\frac{13.5}{5.83}}$

$d_2=0.87 m$

$d_2=0.87\time 100=87 cm$

1 m=100 cm

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3 years ago

What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of −25.0 nc? g?

Daniel [21]

To calculate the force between two negative charges, we use the formula which is given by the Coulomb`s Law as

$F=\frac{kq_{1}q_{2} }{r^{2} }$

Here, $q_{1}$ and $q_{2}$ are the charges on the pith balls, r is the separation between the charges and k is constant and its value is $8.99\times 10^9 N m^2/C^2$ .

Given $q_{1} =q_{2} =-25 nC=-25\times 10^{-9}C$ and $r=8 cm=8\times10^{-2} m$ .

Substituting these values in above formula we get,

$F= 8.99\times 10^9 N m^2/C^2\frac{(-25\times 10^{-9}C)(-25\times 10^{-9}C)}{(8\times10^{-2} m)^2} \\\\\ F= 877929.7\times 10^{-9} N\\\\F=8.8\times10^{-4}N$