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Archy [21]
3 years ago
15

Find the intensity in decibels [i(db)] for each value of i. normal conversation: i = 106i0 i(db) = power saw a 3 feet: i = 1011i

0 i(db) = jet engine at 100 feet: i = 1018i0 i(db) =
Physics
2 answers:
White raven [17]3 years ago
8 0

Answer:

Normal Conversation: i=106i0

i(dB)=60

Power saw a 3 feet: i=1011i0

i(dB)=110

Jet engine at 100 feet: i=1018i0

i(dB)=180

Explanation:

if these are the same as edge, then these are the answers! :)

zalisa [80]3 years ago
6 0

Answer:

1. 60

2. 110

3. 180

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v = ir

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Here, current (i) is measured by Ammeter which should be connected in series of any electrical circuit.

voltage (v) is measured by Voltmeter which should be connected parallel to the external resistance (r).

In the given experiment, the first arrangement of the circuit will show the smallest error because the voltmeter is connected exactly parallel to the external resistance.

In the second arrangement, the voltmeter is connected across external resistance (r) and Ammeter (A) and in this case, the voltmeter will not measure the exact potential drop across the external resistance (r). So, there would be more error.

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3 years ago
a student throws three pencils into the trash can, only one lands in the can and the other hits the window each time? precise or
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1 mole of air undergoes a Carnot cycle. The hot reservoir is at 800 oC and the cold reservoir is at 25 oC. The pressure ranges b
BabaBlast [244]

Answer:

The net work produced is 30.37 KJ

The efficiency of cycle is 72.3%

Explanation:

For the net work produced, we have the formula:

Work = (Th - Tc)(Sh - Sc)(M)

Where,

Th = higher temperature = 800° C + 273 = 1073 k

Tc = lower temperature = 25° C + 273 = 298 k

Sh = specific entropy at higher temperature

Sc = specific entropy at lower temperature

M = molar mass of air

Using, ideal gas table to find entropy. The table is attached.

therefore,

Work = (1073 k - 298 k)(3.0485235 KJ/kg.k - 1.69528 KJ/kg.k)(0.02896 kg)

Work = (775 k)(1.3532435 KJ/kg.k)(0.02896 kg)

<u>Work = 30.37 KJ</u>

Now, for the efficiency (n), we have a formula:

n = 1 - Tc/Th

n = 1 - (298 k)/(1073 k)

<u>n = 0.723 = 72.3 %</u>

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