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Hatshy [7]
3 years ago
7

The first dancer in the line is 10 m from the speaker playing the music; the last dancer in the line is 120 m from the speaker.

Approximately how much time elapses between when the sound reaches the nearest dancer and when it reaches the farthest dancer
Physics
1 answer:
abruzzese [7]3 years ago
3 0

Answer: 0.321 seconds

Explanation:

Let assume that air has a temperature of 20 °C. Sound speed at given temperature is 343 \frac{m}{s}. As sound spreads at constant speed, time can be easily found by using this formula:

\Delta t = \Delta t_{far} - \Delta t_{near}

\Delta t = \frac{x_{far}-x_{near}}{v_{sound,air}}

\Delta{t} = \frac{120 m - 10 m}{343 \frac{m}{s} }\\\\\Delta {t} = 0.321 sec

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3 years ago
What is the kinetic energy of a 10-kg bicycle moving at 10 m/s?​
Stolb23 [73]

Answer:

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K.E

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3 years ago
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inysia [295]

Answer:

Best explains Jamming

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<em>The deliberate radiation of electromagnetic (EM) energy to degrade or neutralize the radio frequency long-haul supervisory control and data acquisition (SCADA) communications links, best explains what?</em>

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6 0
3 years ago
what frequency of light must an electron in hydrogen absorb to jump from the n=2 state to the n=5 state? (a) 2.86 Hz (b) 4.08 Hz
Ostrovityanka [42]

Answer:

(c) 6.91x10^14 Hz

Explanation:

Find the level energy of n=2 and n=5, using  the formula:

E = -E_0/n^2

where  E_0=13.6eV

E_2 =\frac{-13.6}{2^2}=-3.4eV

E_5 =\frac{-13.6}{5^2}=-0.544eV

To jump from n=2 to n=5 the electron absorbs a photon with energy equal to (-0.544) - (-3.4)=2.856eV, using the next formula to find specific wavelength \lambda to that energy

E = hc/\lambda

Where c is the speed of light (c=3 \times10^8m/s) and h is Planck's constant (h=4.14\times10^{-15}eVs). Solve for \lambda:

E = hc/\lambda\\\lambda E = hc\\\lambda = \frac{hc}{E} \\\lambda = \frac{(4.14\times10^{-15})(3 \times10^8)}{2.856}=4.35\times10^{-7}m

The frequency of this wavelength is calculated with this formula:

f=\frac{c}{\lambda}

f=\frac{3\times10^8}{4.3487\times10^{-7}} =6.89\times10^{14}Hz\approx6.9\times10^{14}Hz

8 0
4 years ago
Cathode-ray tubes produce images on the principle of induced emf. true or false?
yan [13]
The appropriate response is false. Cathode-beam ray does not deliver pictures on the guideline of instigated emf. The cathode-ray is a high-vacuum tube in which cathode beams deliver an iridescent picture on a fluorescent screen, utilized mostly in TVs and workstations.
3 0
3 years ago
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