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3 years ago

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Two aluminized optical flats 15 cm in diameter are separated by a gap of 0.04 mm, forming a capacitor. what is the capacitance i

n picofarads

1 answer:

Nikitich [7]3 years ago

6 0

Answer: 3978 pF

Explanation:

Capacitance can be defined as ability to store charge. It is the ratio of charge over electric potential.

$C =\epsilon_o \frac{A}{d}$

where, A is the area of the plates and d is the distance between plates forming the capacitor.

diameter of the plates, di = 15 cm =0.15 m

radius of the plate, r = di/2 = 0.075 m

Area of the plates, A = πr² = 0.018 m²

Distance between the plates, d = 0.04 mm = 0.00004 m

Permittivity, ∈₀ = 8.84 × 10⁻¹² F/m

Capacitance, $C = 8.84 \times 10^{-12} F/m \times \frac{0.018 m^2}{0.00004 m} = 3978 \times 10^{-12} F = 3978 pF$

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When the magnetic flux through a single loop of wire increases by , an average current of 40 A is induced in the wire. Assuming

Zielflug [23.3K]

COMPLETE QUESTION:

<em>When the magnetic flux through a single loop of wire increases by </em>30 Tm^2<em> , an average current of 40 A is induced in the wire. Assuming that the wire has a resistance of </em><em>2.5 ohms </em><em>, (a) over what period of time did the flux increase? (b) If the current had been only 20 A, how long would the flux increase have taken?</em>

Answer:

(a). The time period is 0.3s.

(b). The time period is 0.6s.

Explanation:

Faraday's law says that for one loop of wire the emf $\varepsilon$ is

$(1). \: \: \varepsilon = \dfrac{\Delta \Phi_B}{\Delta t }$

and since from Ohm's law

$\varepsilon = IR$ ,

then equation (1) becomes

$(2). \: \:IR= \dfrac{\Delta \Phi_B}{\Delta t }.$

(a).

We are told that the change in magnetic flux is $\Phi_B = 30Tm^2$ , the current induced is $I = 40A$ , and the resistance of the wire is $R = 2.5\Omega$ ; therefore, equation (2) gives

$(40A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },$

which we solve for $\Delta t$ to get:

$\Delta t = \dfrac{30Tm^2}{(40A)(2.5\Omega)},$

$\boxed{\Delta t = 0.3s}$ ,

which is the period of time over which the magnetic flux increased.

(b).

Now, if the current had been $I =20A$ , then equation (2) would give

$(20A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },$

$\Delta t = \dfrac{30Tm^2}{(20A)(2.5\Omega)},$

$\boxed{\Delta t = 0.6 s\\}$

which is a longer time interval than what we got in part a, which is understandable because in part a the rate of change of flux $\dfrac{\Delta \Phi_B}{\Delta t}$ is greater than in part b, and therefore , the current in (a) is greater than in (b).

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3 years ago

A baseball is thrown 50.0m in 3.0s, what is the average speed of the baseball? Calculated answer is 16.666667! What is the corre

castortr0y [4]

Answer:

20m/s

Explanation:

Given parameters:

Distance of throw = 50m

Time = 3s

Unknown:

Average speed = ?

Solution:

Average speed is distance divided by time;

Average speed = $\frac{distance }{time}$

Insert the parameters and solve;

Average speed = $\frac{50}{3}$ = 16.66667m/s = 20m/s

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3 years ago

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GuDViN [60]

Explanation:

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3 years ago

A sample of an ideal gas has a volume of 2.37 L at 2.80×102 K and 1.15 atm. Calculate the pressure when the volume is 1.68 L and

iogann1982 [59]

Answer:

$p_2 = 1.76 atm$

Explanation:

given data:

v_1 = 2.37 L

v_2 = 1.68 L

p_1 =1.15 atm

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t_1 = 280 K

t_2 = 304 K

from Gas Law Equation

, WE HAVE

$\frac{p_1 v_1}{t_1} =\frac{p_2 v_2}{t_2}$

Putting the values

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$9.733*10^{-3} = \frac{p_2 *1.68}{304}$

$9.733*10^{-3}*304 = p_2*1.68$

$\frac{9.733*10^{-3}*304}{1.68} =p_2$

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4 years ago

A crane raises a 12,000 N marble sculpture at a constant velocity onto a pedestal 1.5 m above the ground outside an art museum.

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For the work, applicate formula:

$\boxed{\boxed{\green{\bf{W = F\times d}}}}$

According our data:

Replacing:

W = 12000 N * 1,5 m

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3 years ago