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ikadub [295]
3 years ago
6

) A steel guitar string with a diameter of 1.00 mm is stretched between supports 80.0 cm apart. The temperature is 0.0°C. (a) Fi

nd the mass per unit length of this string. (Use the value 7.86 x103 for the density (b) The fundamental frequency of transverse oscillations of the string is 200 Hz. What is the tension in the string? (c) If the temperature is raised to 30.0°C, find the resulting values of the tension and the fundamental frequency. Assume that both the Young’s modulus and the average coefficient of expansion have constant values between 0.0°C and 30.0°C.

Physics
1 answer:
ladessa [460]3 years ago
7 0

Answer: a. Mass per unit length =0.0245kg/m

b. Tension =2.45x10^-8N

C. Tension = 2.45 x10^-8N

Fundamental frequency =200Hz

Explanation:

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Which is the best example of the use of imagery in a sentence?
Talja [164]

Answer:

Common Examples of Imagery

Taste: The familiar tang of his grandmother's cranberry sauce reminded him of his youth. Sound: The concert was so loud that her ears rang for days afterward. Sight: The sunset was the most gorgeous they'd ever seen; the clouds were edged with pink and gold.

I hope it's helpful!

6 0
3 years ago
1. A listener stands 20.0 m from a speaker that pumps out music with a power output of 100.0 W.
marta [7]

(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².

(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².

(1.c) The relative intensity of the sound as heard by the listener is 103 dB.

(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.

(2.b) The frequency of the sound heard by the suspect is 614.3 Hz.

<h3>Surface area being vibrated</h3>

The surface area being vibrated by the time the sound reaches the listener is calculated as follows;

A = 4πr²

A = 4π x (20)²

A = 5,026.55 m²

<h3>Intensity of the sound</h3>

The intensity of the sound is calculated as follows;

I = P/A

I = (100) / (5,026.55)

I = 0.02 W/m²

<h3>Relative intensity of the sound</h3>

B = 10log(\frac{I}{I_0} )\\\\B = 10 \times log(\frac{0.02}{10^{-12}} )\\\\B = 103 \ dB

<h3>Speed of sound at the given temperature</h3>

v= 331.3\sqrt{1 + \frac{T}{273} } \\\\v = 331.3\sqrt{1 + \frac{15}{273} } \\\\v = 340.3 \ m/s

<h3>Frequency of the sound</h3>

The frequency of the sound heard is determined by applying Doppler effect.

f_o = f_s(\frac{v \pm v_0}{v \pm v_s} )

where;

  • -v₀ is velocity of the observer moving away from the source
  • -vs is the velocity of the source moving towards the observer
  • fs is the source frequency
  • fo is the observed frequency
  • v is speed of sound

f_0 = f_s(\frac{v-v_0}{v- v_s} )

f_0 = 512(\frac{340.3 - 10}{340.3 - 65} )\\\\f_0 = 614.3 \ Hz

Learn more about intensity of sound here: brainly.com/question/17062836

3 0
2 years ago
What is one way that early scientific practice differed from modern scientific practice?
mote1985 [20]
Early hypotheses were not based on observations.
Early hypotheses were not tested by experimentation.
Early hypotheses were formed from scientific questions.
Early hypotheses were influenced by creative thinking
6 0
3 years ago
Read 2 more answers
Two point charges each have a value of 3.0 c and are separated by a distance of 4.0 m. what is the electric field at a point mid
swat32
 <span>Place a test charge in the middle. It is 2cm away from each charge. 
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point. 
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out. 
THIS IS A TRICK QUESTION. 
THe electric field exactly midway between them = 0/Q = 0. 
But if the point moves even slightly you need the following formula 
F= (1/4Piε)(Q1Q2/D^2) 
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)</span>
4 0
3 years ago
Suppose you pour 0.250 kg of 20.0°C water into a 0.600 kg aluminum pan off the stove with a temperature of 173°C. Assume that th
lapo4ka [179]

Answer:

T_f=5.0116^{\circ}C

Explanation:

Given:

  • mass of water, m_w=0.25\ kg
  • initial temperature of water, T_i_w=20^{\circ}C
  • initial temperature of pan, T_i_p=173^{\circ}C
  • mass of pan, m_p=0.6\ kg
  • mass of water evapourated, m_v=0.03\ kg
  • specific heat of water, c_w=4186\ J.kg^{-1}.K^{-1}
  • specific heat of aluminium pan, c_a=900\ J.kg^{-1}.K^{-1}
  • latent heat of vapourization, L=2256000\ J.kg^{-1}

<u>Using the equation of heat:</u>

<em>Here, initially certain mass of water is vapourised first and then the remaining mass of water comes in thermal equilibrium with the pan.</em>

m_p.c_a.(T_{ip}-T_f)=m_v.L+(m_w-m_v).c_w.(T_f-T_{iw})

0.6\times 900\times (173-T_f)=0.03\times 2256000+(0.25-0.03)\times 4186\times (T_f-20)

T_f=5.0116^{\circ}C

5 0
3 years ago
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