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oksian1 [2.3K]
4 years ago
12

Juan lives 100 m away from Bill/ What is Juan's average speed if he reaches Bill's home in 50 s?

Physics
1 answer:
yanalaym [24]4 years ago
3 0
Data:
ΔS (Total Distance Covered<span>) = 100 m
</span>ΔT (Total time taken<span>) = 50 s
Savg (</span><span>Average Speed) = ? (m/s)
</span><span>
Formula:
</span>S_{Avg} =  \frac{\Delta\:S}{\Delta\:T}<span>

Solving:
</span>S_{Avg} = \frac{\Delta\:S}{\Delta\:T}
S_{Avg} = \frac{100}{50}
\boxed{\boxed{S_{Avg} =2\:m/s}}\end{array}}\qquad\quad\checkmark<span>

</span>

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A circular parallel plate capacitor is constructed with a radius of 0.52 mm, a plate separation of 0.013 mm, and filled with an
olya-2409 [2.1K]

Answer:

289282

Explanation:

r = Radius of plate = 0.52 mm

d = Plate separation = 0.013 mm

A = Area = \pi r^2

V = Potential applied = 2 mV

k = Dielectric constant = 40

\epsilon_0 = Electric constant = 8.854\times 10^{-12}\ \text{F/m}

Capacitance is given by

C=\dfrac{k\epsilon_0A}{d}

Charge is given by

Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}

Number of electron is given by

n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{4.6285\times 10^{-14}}{1.6\times10^{-19}}\\\Rightarrow n=289281.25\ \text{electrons}

The number of charge carriers that will accumulate on this capacitor is approximately 289282.

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