A generalized ____________ contains an antenna complex and a reaction center.

The ultimate source of energy that powers the Sun is__________.

dolphi86 [110]

Answer:C

Explanation:

Mass energy of hydrogen fusing into helium

5 0

1 year ago

Read 2 more answers

Two workhorses tow a barge along a straight canal. Each horse exerts a constant force of magnitude F, and the tow ropes make an

morpeh [17]

Answer:

a) Fvt cosθ

b) Fv cosθ

Explanation:

Each horse exerts a force = F

the rope is inclined at an angle = θ

speed of each horse = v

a) In time t, the distance traveled d = speed x time

i.e d = v x t = vt

also, the resultant force = F cosθ

Work done W = force x distance

W = F cosθ x vt = Fvt cosθ

b) Power provided by the horse P = force x speed

P = F cosθ x v

P = Fv cosθ

8 0

1 year ago

To persuade my audience that the use of mobile communication devices by drivers—even when they are hands-free—is contributing to

pickupchik [31]

Answer:

specific purpose statement

Explanation:

It is a specific purpose statement made for a persuasive speech on the question of fact.

A specific purpose statement helps to build on the general purpose (that is to inform) and to make it more specific to the audience. So if the first speech is an informative speech, our general purpose is to inform our audience about a very specific realm of knowledge.

A specific purpose statement is given to audience to persuade on specific information.

8 0

2 years ago

he first excited state of the helium atom lies at an energy 19.82 eV above the ground state. If this excited state is three-fold

bekas [8.4K]

Answer:

Relative population is 2.94 x 10⁻¹⁰.

Explanation:

Let N₁ and N₂ be the number of atoms at ground and first excited state of helium respectively and E₁ and E₂ be the ground and first excited state energy of helium respectively.

The ratio of population of atoms as a function of energy and temperature is known as Boltzmann Equation. The equation is:

$\frac{N_{1} }{N_{2} }$ = $\frac{g_{1}e^{\frac{-E_{1} }{KT} } }{g_{2}e^{\frac{-E_{2} }{KT} }}$

$\frac{N_{1} }{N_{2} }$ = $\frac{g_{1}e^{\frac{-(E_{1}-E_{2}) }{KT} } }{g_{2}}$

Here g₁ and g₂ be the degeneracy at two levels, K is Boltzmann constant and T is equilibrium temperature.

Put 1 for g₁, 3 for g₂, -19.82 ev for (E₁ - E₂) and 8.6x10⁵ ev/K for K and 10000 k for T in the above equation.

$\frac{N_{1} }{N_{2} }$ = $\frac{1\times e^{\frac{-(-19.82)}{8.6\times 10^{-5}\times 10000} } }{3}$

$\frac{N_{1} }{N_{2} }$ = 3.4 x 10⁹

$\frac{N_{2} }{N_{1} }$ = 2.94 x 10⁻¹⁰

5 0

2 years 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

beam is a Gaussian beam with the waist located at the output. And the beam

waist size is approximately the x-y dimensions of the cavity.)

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

= ×

××× == −

−

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