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11111nata11111 [884]
2 years ago
8

Two identical conducting spheres, A and B, sit atop insulating stands. When they are touched, 1.51 × 1013 electrons flow from sp

here B to sphere A. If the total net charge on the spheres is +3.68 μC, what was the initial charge on sphere B?
Physics
1 answer:
erma4kov [3.2K]2 years ago
6 0

Answer:

A = -0.576 μC

B = 4.256 μC

Explanation:

Suppose a single electron charge is 1.6\times10^{-19}C. Then the total charge that is flowing from B to A is:

1.6\times10^{-19} * 1.51 \times 10^{13} = 2.416\times10^{-6}C = 2.416 \mu C

Let A and B be the initial charge of spheres A and B, respectively. Since the net charge is 3.68μC we have the following equation

A + B = 3.68 (1)

When they touch 2.416μC flows from B to A, then they are equal, so we have the following equation

A + 2.416 = B - 2.416

-A + B = 2.416 + 2.416 = 4.832 (2)

Add equation (1) to equation (2) we have

2B = 3.68 + 4.832 = 8.512

B = 8.512 / 2 = 4.256 \mu C

A = 3.68 - B = 3.68 - 4.256 = -0.576 \mu C

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A 1500 W radiant heater is constructed to operate at 115 V. (a) What will be the current in the heater? (b) What is the resistan
OlgaM077 [116]

Answer:

a) I = 13.04 A

b)  R = 8.82 ohms

c) 1291.87 kilocalories are generated an hour.

Explanation:

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a) we know that:

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R = V/I

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c) the number of kilocalories generated in one hour by the heater is just the energy the heater produces in one hour which is given by:

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since 1 calorie = 4.81 J

1 kilocalorie = 0.001 calories

E = 5400000/4.18 ≈ 1291866.029 calories ≈1291.87 kilocalories

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1 year ago
Consider an optical cavity of length 40 cm. Assume the refractive index is 1, and use the formula for Icavity vs wavelength to p
Bad White [126]

Answer:

Diode Lasers  

Consider a InGaAsP-InP laser diode which has an optical cavity of length 250  

microns. The peak radiation is at 1550 nm and the refractive index of InGaAsP is  

4. The optical gain bandwidth (as measured between half intensity points) will  

normally depend on the pumping current (diode current) but for this problem  

assume that it is 2 nm.  

(a) What is the mode integer m of the peak radiation?  

(b) What is the separation between the modes of the cavity? Please express your  

answer as Δλ.  

(c) How many modes are within the gain band of the laser?  

(d) What is the reflection coefficient and reflectance at the ends of the optical  

cavity (faces of the InGaAsP crystal)?  

(e) The beam divergence full angles are 20° in y-direction and 5° in x-direction  

respectively. Estimate the x and y dimensions of the laser cavity. (Assume the  

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Solution:  

(a) The wavelength λ of a cavity mode and length L are related by  

n

mL

2

λ = , where m is the mode number, and n is the refractive index.  

So the mode integer of the peak radiation is  

1290

1055.1

10250422

6

6

= ×

××× == −

−

λ

nL

m .  

(b) The mode spacing is given by nL

c f 2

=Δ . As

λ

c f = , λ

λ

Δ−=Δ 2

c f .  

Therefore, we have nm

nL f

c

20.1

)10250(42

)1055.1(

2 || 6

2 2 26

= ×××

× ==Δ=Δ −

− λλ λ .  

(c) Since the optical gain bandwidth is 2nm and the mode spacing is 1.2nm, the  

bandwidth could fit in two possible modes.  

For mode integer of 1290, nm

m

nL 39.1550

1290

10250422 6

= ××× ==

−

λ

Take m = 1291, nm

m

nL 18.1549

1291

10250422 6

= ××× ==

−

λ

Or take m = 1289, nm

m

nL 59.1551

1289

10250422 6

= ××× ==

−

λ .

Explanation:

8 0
2 years ago
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