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PtichkaEL [24]
3 years ago
14

Consider a blackbody that radiates with an intensity i1 at a room temperature of 300k. At what intensity i2 will this blackbody

radiate when it is at a temperature of 400k?
Physics
1 answer:
alexgriva [62]3 years ago
4 0

Answer:

3.16 i1

Explanation:

T1 = 300 k

I1 = i1

T2 = 400 k

i2 = ?

The intensity is directly proportional to the four powers of the absolute temperature.

I ∝ T^4

So, \frac{i_{2}}{i_{1}}=\frac{T_{2}^{4}}{T_{1}^{4}}

\frac{i_{2}}{i_{1}}=\frac{400^{4}}{300^{4}}

i2 = 3.16 i1

Thus, the intensity becomes 3.16 i1.

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Light energy with very high amplitude would be perceived as ________.
s2008m [1.1K]

Answer:

High amplitude is equivalent to loud sounds.

Explanation:

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3 years ago
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Mr. MacDougall got his vehicle stuck in the snow. Being the nice student that you are, you stop to help Mr. MacDougall out of th
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Explanation:

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4 0
2 years ago
The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde
Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

S_1 S_2=3m

S_1 O=4m

By Pythagoras theorem

                                 S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m

Now

The intensity at O when both speakers are on is given by

I=4I_1 cos^2(\pi \frac{\delta}{\lambda})

Here

  • I is the intensity at O when both speakers are on which is given as 6 W/m^2
  • I1 is the intensity of one speaker on which is 6  W/m^2
  • δ is the Path difference which is given as

                                           \delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m

  • λ is wavelength which is given as

                                             \lambda=\frac{v}{f}

      Here

              v is the speed of sound which is 320 m/s.

              f is the frequency of the sound which is to be calculated.

                                  16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )

where k=0,1,2

for minimum frequency f_1, k=1

                                  {f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz

So the minimum frequency is 702.22 Hz

5 0
3 years ago
Two people walking on a sidewalk have the following
vova2212 [387]

Answer:

X2 is fasteer

x=0 will go to Xi

Explanation:

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3 years ago
In a coiled spring, the particles of the medium vibrate to and fro about their
Kitty [74]

Answer:

In a coiled spring, the particles of the medium vibrate to and fro about their mean positions at an angle of

A. 0° to the direction of propagation of wave

Explanation:

The waveform of a coiled spring is a longitudinal wave, which is made up of vibrations of the spring which are in the same direction as the direction of the wave's advancement

As the coiled spring experiences a compression force and is then released, it experiences a sequential movement of the wave of the compression that extends the length of the coiled spring which is then followed by a stretched section of the coiled spring in a repeatedly such that the direction of vibration of particles of the coiled is parallel to direction of motion of the wave

From which we have that the angle between the direction of vibration of the particles of the coiled spring and the direction of propagation of the wave is 0°.

8 0
2 years ago
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