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2 years ago

9

An object is 6.0 cm in front of a converging lens with a focal length of 10 cm.Use ray tracing to determine the location of the

image. in cm

1 answer:

Masja [62]2 years ago

8 0

As we know that

$\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}$

here we know that

$d_0 = 6 cm$

$f = 10 cm$

now from above equation we have

$\frac{1}{d_i} + \frac{1}{6} = \frac{1}{10}$

$d_i = -15 cm$

so image will form on left side of lens at a distance of 15 cm

This image will be magnified and virtual image

Ray diagram is attached below here

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A p-type Si sample is used in the Haynes-Shockley experiment. The length of the sample is 2 cm, and two probes are separated by

Airida [17]

Answer:

Mobility of the minority carriers, $\mu_{n} =1184.21 cm^{2} /V-sec$

Diffusion coefficient for minority carriers, $D_{n} = 29.20 cm^2 /s$

Verified from Einstein relation as $\frac{D_{n} }{\mu_{n} } = 25 mV$

Explanation:

Length of sample, $l_{s} = 2 cm$

Separation between the two probes, L = 1.8 cm

Drift time, $t_{d} = 0.608 ms$

Applied voltage, V = 5 V

Mobility of the minority carriers ( electrons), $\mu_{n} = \frac{V_{d} }{E}$

Where the drift velocity, $V_{d} = \frac{L}{t_{d} }$

$V_{d} = \frac{1.8}{0.608 * 10^{-3} } \\V_{d} = 2960.53 cm/s$

and the Electric field strength, $E = \frac{V}{l_{s} }$

E = 5/2

E = 2.5 V/cm

Mobility of the minority carriers:

$\mu_{n} = 2960.53/2.5\\\mu_{n} =1184.21 cm^{2} /V-sec$

The electron diffusion coefficient, $D_{n} = \frac{(\triangle x)^{2} }{16 t_{d} }$

$\triangle x = (\triangle t )V_{d}$ , where Δt = separation of pulse seen in an oscilloscope in time( it should be in micro second range)

$\triangle x = \frac{(\triangle t) L}{t_{d} } \\\triangle x = \frac{180*10^{-6} * 1.8}{0.608*10^{-3} }\\\triangle x =0.533 cm$

$D_{n} = \frac{0.533^{2} }{16 * 0.608 * 10^{-3} }\\D_{n} = 29.20 cm^2 /s$

For the Einstein equation to be satisfied, $\frac{D_{n} }{\mu_{n} } = \frac{KT}{q} = 0.025 V$

$\frac{D_{n} }{\mu_{n} } = \frac{29.20}{1184.21} \\\frac{D_{n} }{\mu_{n} } = 0.025 = 25 mV$

Verified.

4 0

2 years ago

Two bar magnets are labeled A and B. The ends of each magnet are numbered 1 or 2, but the poles are not labeled. When A1 is brou

vichka [17]

It is definitely letter D. <span>A1 and B1 are like poles, but there is not enough information to tell whether they are north poles or south poles.

A1 and B1 is either both north poles or both south poles. Repulsion of both magnets says it all--like poles always repel while opposite poles always attract. Thus, the best conclusion to this would be choice D.</span>

3 0

2 years ago

Read 2 more answers

The value of (997)1/3 according to binomial theorem is

postnew [5]

Answer:

9.99

Explanation:

The value of (997)^1/3

(997)^1/3

997 = (1000 - 3)

(1000 - 3)^1/3

Expanding :

[1000(1 - 3/1000)]^1/3

1000^1/3 * (1 - 3/1000)^1/3

Cube root of 1000

10 * (1 - 3/1000 * 1/3)

10 * (1 - 1/1000)

10 * (1 - 0.001)

10(0.999)

= 9.99

Hence, the value of (997)^1/3 according to binomial theorem is 9.99

5 0

1 year ago

Sea breezes that occur near the shore are attributed to a difference between land and water with respect to what property?

ddd [48]

Answer:

a. mass density

Explanation:

<em>Land and sea breeze that occur near the shore are due to the variation of mass density of air with change in temperature.</em>

When the air gets heated it becomes rarer in density and thus rises up in the atmosphere and its space is occupied by a cooler and denser air that flows to the place.

<em>During the day the land is warmer than the sea so the sea breeze blows and during the night the water bodies are warmer than the land so the land breeze blows.</em>

7 0

2 years ago

A semi is traveling down the highway at a velocity of v = 26 m/s. The driver observes a wreck ahead, locks his brakes, and begin

Dovator [93]

Answer:

fcosθ + Fbcosθ =Wtanθ

Explanation:

Consider the diagram shown in attachment

fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)

Fbx= Fbcosθ ( Fbx: component of braking force in x-direction ; Fb: braking force)

Wx= Wtanθ (Wx: component of weight in x-direction ; W: Weight of semi)

sum of x-direction forces = 0

fx+ Fbx=Wx

fcosθ + Fbcosθ =Wtanθ

7 0

1 year ago