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Jobisdone [24]
3 years ago
15

The sound level at a distance of 2.30 m from a source is 115 dB. At what distance will the sound level have the following values

? (a) 100 dB Your response differs from the correct answer by more than 100%. m (b) 10.0 dB
Physics
1 answer:
Aleksandr [31]3 years ago
5 0

Answer:

distance is 13 m for 100 dB

distance is 409 km for 10 dB

Explanation:

Given data

distance r = 2.30 m

source β = 115 dB

to find out

distance at sound level 100 dB and 10 dB

solution

first we calculate here power and intensity and with this power and intensity we will find distance

we know sound level  β  = 10 log(I/I_{0})        ......................a

put here value (I/I_{0}) = 10^−12 W/m² and  β = 115

115  = 10 log(I/10^−12)

so

I = 0.316228 W/m²

and we know power = intensity × 4π r²    ...............b

power = 0.316228 × 4π (2.30)²

power = 21.021604 W

we know at 100 dB intensity is 0.01 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 0.01 × 4π r²

so by solving r

r = 12.933855 m    = 13 m

distance is 13 m

and

at 10 dB intensity is 1 × 10^–11 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 1 × 10^–11 × 4π r²

by solving r we get

r = 409004.412465 m = 409 km

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An intravenous (IV) system is supplying saline solution to a patient at the rate of 0.06 cm3/s through a needle of radius 0.2 mm
horsena [70]

Answer:

Pressure applied to the needle is 7528 Pa

Explanation:

As we know by poiseuille's law of flow of liquid through a cylindrical pipe

the rate of flow through the pipe is given as

Q = \frac{\Delta P \pi r^4}{8\eta L}

now we know that

Q = 0.06 \times 10^{-6} m^3/s

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\eta = 1\times 10^{-3} Pa s

now we have

6 \times 10^{-8} = \frac{\Delta P \pi (0.2 \times 10^{-3})^4}{8(1 \times 10^{-3})6.32 \times 10^{-2}}

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8 0
3 years ago
you have been called to testify as a as an expert witness in a trial involving a head-on collision Car A weighs 1515 pounds and
GrogVix [38]

Answer:

70.6 mph

Explanation:

Car A mass= 1515 lb

Car B mass=1125 lb  

Speed of car B is 46 miles/h

Distance before locking, d=19.5 ft

Coefficient of kinetic friction is 0.75

Initial momentum of car B=mv where m is mass and v is velocity in ft/s  

46 mph*1.46667=67.4666668 ft/s

Momentum_B=1125*67.4666668 ft/s

Initial momentum of car A is given by

Momentum_A=1515v_a where v_a is velocity of A

Taking East as positive and west as negative then the sum of initial momentum is

1515v_a-(1125*67.4666668 ft/s)

The common velocity is represented as v_c hence after collision, the final momentum is

Momentum_final=(m_a+m_b)v_c=(1515+1125)v_c=2640v_c

From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence

1515v_a-(1125*67.4666668 ft/s)= 2640v_c

The acceleration of two cars a=-\mug=-0.75*32.17=-24.1275 ft/s^{2}

From kinematic equation

v^{2}=u^{2}+2as hence

v^{2}-u^{2}=2as

0^{2}-(v_c)^{2}=2*-24.1275*19.5

v_c=\sqrt{2*24.1275*19.5}=30.67 ft/s

Substituting the value of v_c in equation 1515v_a-(1125*67.4666668 ft/s)= 2640v_c

1515v_a-(1125*67.4666668 ft/s)= 2640*30.67

1515v_a=(1125*67.4666668 ft/s)+2640*30.67

v_a=\frac {156868.8}{1515}=103.5438 ft/s

\frac {103.5438}{1.46667}=70.59787 mph\approx 70.60 mph

3 0
3 years ago
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