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1 year ago

An air-filled 20-μf capacitor has a charge of 60 μc on its plates. how much energy is stored in this capacitor?

Physics

1 answer:

Triss [41]1 year ago

3 0

Thew energy stored in a capacitor of capacitance $C$ and voltage between the plates $V$ is

$E=\frac{1}{2} CV^2=\frac{1}{2C} Q^2$ .

Substituting numerical value

$E=\frac{1}{2*20*10^{-6}} (60*10^{-6})^2\\ E=90\; \mu J$

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A thin layer of oil with index of refraction no = 1.47 is floating above the water. The index of refraction of water is nw = 1.3

garik1379 [7]

Answer:

A thin layer of oil with index of refraction ng = 1.47 is floating above the water. The index of refraction of water is nw = 1.3. The index of refraction of air is na= 1. A light with wavelength λ = 775 nm goes in from the air to oil and water.

Part (a) Express the wavelength of the light in the oil, λ₀, in terms of λ and n⁰ (b) Express the minimum thickness of the film that will result in destructive interference, t min, in terms of λ o

(d) Solve for the numerical value of tmin in nm.

Explanation:

n₀ = 1.47

refraction of water = 1.3

refraction of air = 1

wavelength λ = 775 nm

(a) wavelength of light in water ⇒ λ₀ = λ / n₀

(b) minimum thickness of the film that will result in destructive interference

t(min) = λ₀ / 2

t = λ /2n₀

(d) the thickness is

t = 775 / 2(1.47)

= 263.61 nm

4 0

2 years ago

A ball of mass 5.0kg is lifted off the floor a distance of 1.7m. 1. What is the change in the gravitational potential energy of

tangare [24]

Answer:

Explanation:

Change in gravitational energy of the ball = mgh

5 mutiply 10 multiply 1.7 = 85J

Potential energy at height = Kinetic energy at bottom

KE= 85J

Velocity

v=5.83m/s

4 0

1 year ago

If the humidity in a room of volume 450 m3 at 30 ∘C is 75%, what mass of water can still evaporate from an open pan?

Yuliya22 [10]

From tables,

SVP at 30°C = 4.24 kPa

From ideal gas expressions;
n = PV/RT = (4.24*1000*450)/(8.314*303) = 757.4 moles

Now, 75% of 757.4 moles will evaporate leaving 20%. Then, 25% of 757.5 moles...
25% of 757.4 moles = 25/100*757.4 = 189.35 moles
Mass of 189.35 moles = 189.35 moles*18 g/mol = 3408.3 g ≈ 3.4 kg

5 0

2 years ago

A coaxial cable consists of a thin insulated straight wire carrying a current of 2.00 A surrounded by a cylindrical conductor ca

ella [17]

Answer:

B = 15μT

Explanation:

In order to calculate the magnitude of the magnetic field generated by the coaxial cable you use the Ampere's law, which is given by:

$B=\frac{\mu_oI}{2\pi r}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

I: current

r: distance from the wire to the point in which B is calculated

In this case you have two currents with opposite directions, which also generates magnetic opposite magnetic fields. Then, you have (but only for r > radius of the cylindrical conductor) the following equation:

$B_T=B_1-B_2=\frac{\mu_o I_1}{2\pi r}-\frac{\mu_o I_2}{2\pi r}\\\\B_T=\frac{\mu_o}{2\pi r}(I_1-I_2)$ (2)

I1: current of the central wire = 2.00A

I2: current of the cylindrical conductor = 3.50A

r: distance = 2.00 cm = 0.02 m

You replace the values of all parameters in the equation (2), and you use the absolute value because you need the magnitude of B, not its direction.

$|B|=|\frac{4\pi*10^{-7}T/A}{2\pi (0.02m)}(2.00A-3.50A)|=1.5*10^{-5}T\\\\|B|=15*10^{-6}T=15\mu T$

The agnitude of the magnetic field outside the coaxial cable, at a distance of 2.00cm to the center of the cable is 15μT

3 0

2 years ago

A linear accelerator produces a pulsed beam of electrons. The pulse current is 0.50 A, and the pulse duration is 0.10 μs. (a) Ho

Crank

Answer:

a)N = 3.125 * 10¹¹

b) I(avg) = 2.5 × 10⁻⁵A

c)P(avg) = 1250W

d)P = 2.5 × 10⁷W

Explanation:

Given that,

pulse current is 0.50 A

duration of pulse Δt = 0.1 × 10⁻⁶s

a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles

N = Δq/e

charge is given by Δq = IΔt

so,

N = IΔt / e

$N = \frac{(0.5)(0.1 * 10^-^6)}{(1.6 * 10^-^1^9)} \\= 3.125 * 10^1^1$

N = 3.125 * 10¹¹

b) Q = nqt

where q is the charge of 1puse

n = number of pulse

the average current is given as I(avg) = Q/t

I(avg) = nq

I(avg) = nIΔt

= (500)(0.5)(0.1 × 10⁻⁶)

= 2.5 × 10⁻⁵A

C) If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,

eV = K

V = K/e

the power is given by

P = IV

P(avg) = I(avg)K / e

$P(avg) = \frac{(2.5 * 10^-^5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}$

= 1250W

d) Final peak=

P= Ik/e

= $= P(avg) = \frac{(0.5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}\\2.5 * 10^7W$

P = 2.5 × 10⁷W

5 0

2 years ago

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