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Taya2010 [7]
3 years ago
12

You are using a diffraction grating with 2230 lines per centimeter. White light passes through it, and the grating spreads the l

ight out into its spectral components. At what angle does green light of wavelength 508 nm appear in second order
Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
5 0

Answer:

The angle of diffraction is 13.09°

Explanation:

Given:

Wavelength of light \lambda = 508 \times 10^{-9} m

Order of diffraction n = 2

No. lines per centimeter N = 2230

First find the grating element,

  d = \frac{1}{N}

  d = 4.484 \times 10^{-6} m

From the formula of diffraction,

  d \sin \theta = n \lambda

    \sin \theta = \frac{2 \lambda}{d}

    \sin \theta = \frac{2 \times 508 \times 10^{-9} }{4.484 \times 10^{-6} }

    \theta = \sin ^{-1} (0.2265)

    \theta = 13.09°

Therefore, the angle of diffraction is 13.09°

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A 60-W light bulb radiates electromagnetic waves uniformly in all directions. At a distance of 1.0 mm from the bulb, the light i
slega [8]

Answer:

The appropriate solution is:

(a) \frac{1}{4}(I_o)

(b) \frac{1}{4} (u_o)

(c) \frac{1}{2}B_o

Explanation:

According to the question, the value is:

Power of bulb,

= 60 W

Distance,

= 1.0 mm

Now,

(a)

⇒  \frac{I}{I_o} =\frac{r_o_2}{r_2}

On applying cross-multiplication, we get

⇒  I=I_o\times \frac{1_2}{2^2}

⇒     =I_o\times \frac{1}{4}

⇒     =\frac{1}{4} (I_o)

(b)

As we know,

⇒ \frac{u}{u_o} =\frac{I}{I_o}

By putting the values, we get

⇒ u=\frac{1}{4}(u_o)

(c)

⇒ \frac{B^2}{B_o^2} =\frac{u}{u_o}

         =\frac{I}{I_o}

⇒ B=B_o\times \sqrt{\frac{1}{4} }

⇒     =\frac{1}{2}(B_o)

4 0
2 years ago
What humidity reading would be the driest?
Lunna [17]
The weather report never tells the "humidity" of the air.  It tells the
"RELATIVE humidity".  That's the percent of the moisture the air
COULD hold at the current temperature that it's actually holding
right now.

If the air is completely dry, then it's holding NONE of the moisture
that it COULD hold, and the relative humidity is zero percent.

7 0
3 years ago
Assume this 1.20-mm-radius copper wire is electrically neutral in the Earth reference frame, in which it is at rest and carrying
agasfer [191]

Answer:

The charge density in the system is 4.25*10^4C/m

Explanation:

To solve this problem it is necessary to keep in mind the concepts related to current and voltage through the density of electrons in a given area, considering their respective charge.

Our data given correspond to:

r=1*10^{-3}m\\v = 5.2*10^{-4}m/s\\e= 1.6*10^{-19}C

We need to asume here the number of free electrons in a copper conductor, at which is generally of 8.5 *10^{28}m^{-3}

The equation to find the current is

I = VenA

Where

I =Current

V=Velocity

A = Cross-Section Area

e= Charge for a electron

n= Number of free electrons

Then replacing,

I = (5.2*10^{-4})(1.6*10^{-19})(88.5 *10^{28})(\pi(1*10^{-3})^2)

I= 22.11a

Now to find the linear charge density, we know that

I = \frac{Q}{t} \rightarrow Q = It

Where:

I: current intensity

Q: total electric charges

t: time in which electrical charges circulate through the conductor

And also that the velocity is given in proportion with length and time,

V_d = \frac{l}{t} \rightarrow l = V_d t

The charge density is defined as

\lambda = \frac{Q}{l}\\\lambda = \frac{It}{V_d t}\\\lambda = \frac{I}{V_d}

Replacing our values

\lambda = \frac{22.11}{5.20*10{-4}}

\lambda= 4.25*10^4C/m

Therefore the charge density in the system is 4.25*10^4C/m

5 0
3 years ago
Starting fom rest, a car accelerates at a constant rate, reaching 88 km/h in 12 s. (a) What is its acceleration? (b) How far doe
JulsSmile [24]

Answer:

(A)  a=2.0.37m/sec^2

(B) s = 146.664 m

Explanation:

We have given car starts from the rest so initial velocity u = 0 m /sec

Final velocity v = 88 km/hr

We know that 1 km = 1000 m

And 1 hour = 3600 sec

So 88km/hr=88\times \frac{1000}{3600}=24.444m/sec

Time is given t = 12 sec

(A) From first equation of motion v = u+at

So 24.444=0+a\times 12

a=2.0.37m/sec^2

So acceleration of the car will be a=2.0.37m/sec^2

(b) From third equation of motion v^2=u^2+2as

So 24.444^2=0^2+2\times 2.037\times s

s = 146.664 m

Distance traveled by the car in this interval will be 146.664 m

6 0
3 years ago
A heat engine receives heat of 400kW from a heat source of 1100 K, and rejects the waste heat into a heat sink at 320K. It gener
GuDViN [60]

Answer:

Explanation:

efficiency of carnot engine operating between 1100 K and 320 K

= 1100 - 320 / 1100

= .709

efficiency of heat engine = 120 / 400 = .30

2 nd law thermal efficiency of heat engine = .30 x 100  / .709

= 42.31 % .

4 0
2 years ago
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