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3 years ago

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A monochromatic red laser beam emitting 1 mW at a wavelength of 638 nm is incident on a silicon solar cell. Find the following:

a. The number of photons per second incident on the cell b. The maximum possible efficiency of conversion of this laser c. beam to electricity

1 answer:

tia_tia [17]3 years ago

8 0

Answer:

Explanation:

First we calculate the energy of the photon

E=(Planck constant × speed of light in vacuum)÷ wave length

E= $\frac{6.626*10^{34}*2.998*10^{8} }{638*10^{-9} } = 3.114*10^{49}$

Next we find the total energy per second

total energy= $1*10^{-3}W *\frac{1JS^{-1} }{1W} = 1*10^{-3} JS^{-1}$

Next we calculate the number the photon per second

= total energy ÷ energy of 1 photon

= $\frac{1*10^{-3} JS^{-1} }{3.114*10^{49} } = 3.21*10^{-53} \ photons/sec$

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Suppose the farthest distance a person can see without visual aid is 53 cm.

natita [175]

Answer: a. 53cm b. Diverging c. 0.02dioptres

Explanation:

Since the person in question cannot see far object clearly, it shows that the individual is suffering from myopia (short sightedness).

If the object distance (u) is 53cm. for the person to be able to see far way, it means the image must be at infinity. This shows image distance v is infinity. Using the lens formula

1/f = 1/u+1/v

1/f = 1/53 + 1/Infinity

1/f = 1/53 + 0

f=53cm

b) The lens used is concave lens (diverging lens) to diverge all rays outwards

c) Power of a lens P will be 1/53.

P = 0.02dioptres

8 0

3 years ago

A quarter is flipped from a height of 1.45 m above the ground. How much time will it take to reach the ground if the person flip

Artemon [7]

It will take the quarter 0.151 seconds to reach the ground.

<u>Given the following data:</u>

Height = 1.45 meters

Initial velocity = 10.32 m/s

We know that acceleration due to gravity (a) for an object is equal to 9.8 meter per seconds square.

To find how much time it will take the quarter to reach the ground, we would use the second equation of motion.

Mathematically, the second equation of motion is given by the formula;

$S = ut + \frac{1}{2} at^2$

Where:

S is the height or distance covered.

u is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Substituting the values into the formula, we have;

$1.45 = 10.32(t) + \frac{1}{2} (9.8)t^2\\\\1.45 = 10.32t + 4.9t^2\\\\4.9t^2 + 10.32t - 1.45 = 0$

The standard form of a quadratic equation is:

$ax^2 + bx + c = 0$

a = 4.9, b = 10.32 and c = 1.45

We would solve the above quadratic equation by using the quadratic equation formula;

$x = \frac{-b\; \pm \;\sqrt{b^2 - 4ac}}{2a}$

Substituting the values, we have;

$t = \frac{-10.32\; \pm \;\sqrt{10.32^2\; - \;4(4.9)(1.45)}}{2(4.9)}\\\\t = \frac{-10.32\; \pm \;\sqrt{106.5024\; - \;28.42}}{9.8}\\\\t = \frac{-10.32\; \pm \;\sqrt{78.0824}}{9.8}\\\\t = \frac{-10.32\; \pm \;8.84}{9.8}\\\\t = \frac{-10.32\; + \;8.84}{9.8}\\\\t = \frac{1.48}{9.8}$

<em>Time, t = 0.151 seconds.</em>

Therefore, it will take the quarter 0.151 seconds to reach the ground.

Read more: brainly.com/question/8898885

3 0

2 years ago

The reactive force developed by a jet engine to push an airplane forward is called thrust, and the thrust developed by the engin

finlep [7]

The value of the thrust developed by the engine of a boeing 777 in N and Kgf are ;

i) 376616N

ii) 38430Kgf

<h3>What is force?</h3>

Force is a push or a pull. The reactive force always serve to balance the applied force. We are here asked to convert the the thrust developed by the engine of a boeing 777 which is about 85400 lbf to the following units;

i) N

ii)kgf

Thus;

1 Ib = 0.45 Kg

1 lbf = 0.45 Kg * 9.8 m/s^2 = 4.41 N

We know that;

1 lbf = 4.41 N

85400 lbf = 85400 lbf. * 4.41 N/1 lbf

= 376616N

Again;

1 lbf = 0.45 Kgf

85400 lbf = 85400 lbf * 0.45 Kgf/1 lbf

= 38430Kgf

Learn more about force:brainly.com/question/13191643

#SPJ1

3 0

1 year ago

Two carts, one twice as heavy as the other, are at rest on a horizontal track. A person pushes each cart for 8 s. Ignoring frict

GalinKa [24]

Answer:

The correct option is: B that is 1/2 K

Explanation:

Given:

Two carts of different masses, same force were applied for same duration of time.

Mass of the lighter cart = $m$

Mass of the heavier cart = $2m$

We have to find the relationship between their kinetic energy:

Let the KE of cart having mass m be "K".

and KE of cart having mass m be "K1".

As it is given regarding Force and time so we have to bring in picture the concept of momentum Δp and find a relation with KE.

Numerical analysis.

⇒ $KE = \frac{mv^2}{2}$

⇒ $KE = \frac{mv^2}{2}\times \frac{m}{m}$

⇒ $KE = \frac{m^2v^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(mv)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2m}=\frac{(F\times t)^2}{2m}$

Now,

Kinetic energies and their ratios in terms of momentum or impulse.

KE (K) of mass m.

⇒ $K=\frac{(F\times t)^2}{2m}$ ...equation (i)

KE (K1) of mass 2m.

⇒ $K_1=\frac{(F\times t)^2}{2\times 2m}$

⇒ $K_1=\frac{(F\times t)^2}{4m}$ ...equation (ii)

Lets divide K1 and K to find the relationship between the two carts's KE.

⇒ $\frac{K_1}{K} =\frac{(F\times t)^2}{4m} \times \frac{2m}{(F\times t)^2}$

⇒ $\frac{K_1}{K} =\frac{2m}{4m}$

⇒ $\frac{K_1}{K} =\frac{2}{4}$

⇒ $\frac{K_1}{K} =\frac{1}{2}$

⇒ $K_1=\frac{K}{2}$

⇒ $K_1=\frac{1}{2}K$

The kinetic energy of the heavy cart after the push compared to the kinetic energy of the light cart is 1/2 K.

7 0

3 years ago

Hii please help i’ll give brainliest!!

faltersainse [42]

The mass of the box
HOPE THIS HELPS❤️

8 0

3 years ago