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SCORPION-xisa [38]
3 years ago
7

A horse performs 2600 joules of work pulling a wagon for 13 seconds. What is the horse's power? 33800 W 200 W 0.005 W 2600 WA ho

rse performs 2600 joules of work pulling a wagon for 13 seconds. What is the horse's power?
33800 W

200 W

0.005 W

2600 W
Physics
2 answers:
tensa zangetsu [6.8K]3 years ago
6 0

Answer:

200 W

Explanation:

The power is given by the ratio between the amount of work done and the time taken to perform the work:

P=\frac{W}{t}

where in this problem we have

W = 2600 J is the work done

t = 13 s is the time taken to do the work

Substituting the numbers into the equation, we find

P=\frac{2600 J}{13 s}=200 W

OLga [1]3 years ago
5 0

Answer:

200 W

Explanation:

To figure out the power, you need to use the formula: P = work/time<em>. You need to remember that they may ask you for the significant figures, so that is how many digits you have to round by. This question does not specify significant figures.</em>

<em>2600/ 13 =  200 </em>

Your answer is 200 watts

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Answer:

Period of brightness variation and luminosity.

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The Cepheid variables are used as distance indicators. This requires estimation of periods and (usually) intensity-mean magnitudes in order to establish a period—apparent luminosity relation. It is particularly important for the techniques employed to be as accurate and efficient as possible.

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2 years ago
What is the value of n in the balmer series for which the wavelength is 410.2 nm.?
m_a_m_a [10]

The answer is n= 6.

What is Balmer series?

The Balmer series is the portion of the emission spectrum of hydrogen that represents electron transitions from energy levels n > 2 to n = 2. These are four lines in the visible spectrum. They are also known as the Balmer lines. The four visible Balmer lines of hydrogen appear at 410 nm, 434 nm, 486 nm and 656 nm.

For the Balmer series, the final energy level is always n=2. So, the wavelengths 653.6, 486.1, 434.0, and 410.2 nm correspond to n=3, n=4, n=5, and n=6 respectively. Since the last wavelength, 410.2 nm, corresponds to n=6, the next wavelength should logically correspond to n=7.

To solve for the wavelength, calculate the individual energies, E2 and E7, using E=-hR/(n^2). Then, calculate the energy difference between E2 (which is the final) and E7 (which is the initial). Finally, use lamba=hc/E to get the wavelength.

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3 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

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