Sound must travel through a medium/body. Space is a vacuum so there is no medium/body for sound to travel through, so there is no sound.
Answer:
The gplanet is 0.193 m/s^2
Explanation:
The speed of the pulse is:
where
m=mass of the wire=4 g= 4x10^-3 kg
M=mass of the object= 3 kg
Replacing values:
Answer:
660V
Explanation:
V=IR
V=11×60
=660V
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Answer:
124.86 V
Explanation:
We have to first calculate the voltage drop across the copper wire. The copper wire has a length of 358 ft
1 ft = 0.3048 m
358 ft = 109.12 m
The diameter of 2 AWG copper wire (d) = 6.544 mm = 0.006544 m
The area of the wire = πd²/4 = (π × 6.544²)/4 = 33.6 mm²
Resistivity of wire (ρ) = 0.0171 Ω.mm²/m
The resistance of the wire = 
The voltage drop across wire = current * resistance = 6.1 A * 0.056 ohm = 0.34 V
The voltage at end = 125.2 - 0.34 = 124.86 V
There are two units of sound: intensity and in decibels. Decibels are not additive, you must convert it first to units of intensity (W/m²) using this formula:
dB = 10 log(I/10⁻¹²)
A. 100 dB = 10 log(I/10⁻¹²)
Solving for I,
I = 0.01 W/m²
90 dB = 10 log(I/10⁻¹²)
Solving for I,
I = 0.001
Ratio = 0.01/0.001 = 10
<em>Thus,the choir is 10 times more intense than the soloist.</em>
B. Since there are 90 singers, there would be 9 groups of 10-person choir that produces 100 dB or 0.01 W/m². The total intensity would be
Total intensity = 0.01 W/m² (original choir) + 0.001 W/m² (soloist) + 10(0.01 W/m²) (additional 90 singers) = 0.111 W/m²
dB = 10 log(0.111/10⁻¹²) = <em>110.45 dB</em>
C. Rock concert:
120 dB = 10 log(I/10⁻¹²)
Solving for I,
I = 1 W/m²
Ratio = 1/0.111 = 9
<em>Therefore, the rock concert is 9 times more intense than the choir concert.</em>
