How does a pendalum conserve energy

A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$

3 0

3 years ago

A bullet of mass 0.1 kg traveling horizontally at a speed of 100 m/s embeds itself in a block of mass 3 kg that is sitting at re

Xelga [282]

Answer:

(a) the speed of the block after the bullet embeds itself in the block is 3.226 m/s

(b) the kinetic energy of the bullet plus the block before the collision is 500J

(c) the kinetic energy of the bullet plus the block after the collision is 16.13J

Explanation:

Given;

mass of bullet, m₁ = 0.1 kg

initial speed of bullet, u₁ = 100 m/s

mass of block, m₂ = 3 kg

initial speed of block, u₂ = 0

Part (A)

Applying the principle of conservation linear momentum, for inelastic collision;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the block after the bullet embeds itself in the block

(0.1 x 100) + (3 x 0) = v (0.1 + 3)

10 = 3.1v

v = 10/3.1

v = 3.226 m/s

Part (B)

Initial Kinetic energy

Ki = ¹/₂m₁u₁² + ¹/₂m₂u₂²

Ki = ¹/₂(0.1 x 100²) + ¹/₂(3 x 0²)

Ki = 500 + 0

Ki = 500 J

Part (C)

Final kinetic energy

Kf = ¹/₂m₁v² + ¹/₂m₂v²

Kf = ¹/₂v²(m₁ + m₂)

Kf = ¹/₂ x 3.226²(0.1 + 3)

Kf = ¹/₂ x 3.226²(3.1)

Kf = 16.13 J

6 0

3 years ago

Because the top mirror is not perfectly reflective (it reflects 90% of the photons, allowing 10% of them to go through), the pow

allsm [11]

This question is incomplete, the complete question;

you make an interferometer using 50-50 beam splitter and two mirrors, one being a perfect mirror and one which does not reflect all light. The wavelength of the 9 mW incident laser is 400 nm.

Because the top mirror is not perfectly reflective (it reflects 90% of the photons, allowing 10% of them to go through), the power measured at the detector when only the vertical arm is blocked is 2.25 mW, while the power measured at the detector when only the horizontal arm is blocked is only 2.025 mW. Assume initially the intensity is at its maximum. How much would we need to translate the perfect mirror to the right to get a minimum intensity at detector, and what is that minimum intensity

Options;

a) 200 nm; 0.9 mW

b) 100 nm, 0.0059 mW

c) 200 nm; 0 mW

d) 100 nm; 0.9 mW

e) 200 nm; 0.0059 mW

Answer:

the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector and the minimum intensity are;

100 nm; 0.0059 mW

Option b) 100 nm, 0.0059 mW is the correct answer

Explanation:

Given that the instrument here is an interferometer.

Maximum intensity is obtained when the two waves are exactly in phase.

that is the peaks (crusts and troughs) and nodes (zero value points) of the two waves will be at the exact same point when the wave falls on the detector.

The phase factor of this point is taken as ∅ = 0

Now, to get a minimum point, the phase difference between the two waves should be should be ∅ = π

This corresponds to a path difference between the two waves as half of the wavelength. λ/2

The light gets reflected from the mirror.

Hence, when we move the mirror by a length l, the extra/less path the light has to travel is 2l (light is going and coming back)

hence, to get a path difference of λ/2 the mirror should move half of this distance only

so, the mirror should move;

$l$ = λ/4

here, wavelength is 400nm

the length moved by the mirror = 400/4 = 100 nm

The intensity is given by the equation;

$l$ = $l$ 1 + $l$ 2 + 2√ $l$ 1 $l$ 2cos(∅)

where

$l$ 1 = 2.25 mW

$l$ 2 = 2.025 mW

∅ = π

so we substitute

$l$ = 2.25 + 2.025 - 2√(2.25 × 2.025)

$l$ = 4.275 - 4.26907

$l$ = 0.0059

Therefore; the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector and the minimum intensity are;

100 nm; 0.0059 mW

Option b) 100 nm, 0.0059 mW is the correct answer

5 0

3 years ago

A pendulum on a grandfather clock

egoroff_w [7]

4.0 ilynits per second Alaskan es muy du facial in the oscillates 1.99

5 0

3 years ago

A proton is confined within an atomic nucleus of diameter 3.60 fm. part a estimate the smallest range of speeds you might find f

Cerrena [4.2K]

The answer for this problem would be:
Assuming non-relativistic momentum, then you have:
ΔxΔp = mΔxΔv = h / (4)
Δv = h / (4πmΔx)
m ~ 1.67e-27 h ~ 6.62e-34,Δx = 4e-15 -->
Δv ~ 6.62e-34 / (4π * 1.67e-27 * 4e-15) ~ 7,886,270 m/s ~ 7.89e6 m/s
That's about 1% of the speed of light, the assumption that it's non-relativistic.

3 0

3 years ago