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stepladder [879]
3 years ago
11

A popular car stereo has four speakers, each rated at 60 W. In answering the following questions, assume that the speakers produ

ce sound at their maximum power.
A) Find the intensity I of the sound waves produced by one 60-Wspeaker at a distance of 1.0 m.
B) Find the intensity I of the sound waves produced by one 60-Wspeaker at a distance of 1.5 m.
C) Find the intensity I of the sound waves produced by four 60-Wspeakers as heard by the driver. Assume that the driver is located 1.0 m from each of the two front speakers and 1.5 m from each of the two rear speakers.
D)The threshold of hearing is defined as the minimum discernible intensity of the sound. It is approximately 10^(-12) W/m2. Find the distance dfrom the car at which the sound from the stereo can still be discerned. Assume that the windows are rolled down and that each speaker actually produces 0.06 W of sound, as suggested in the last follow-up comment.
Physics
1 answer:
Fantom [35]3 years ago
7 0

Answer:

Explanation:

Intensity of sound = sound energy emitted by source / 4 π d² , where d is distance of the source .

A )

Intensity of sound at 1 m distance = 60 /4 π d²

d = 1 m

Intensity of sound at 1 m distance = 60 /(4 π 1²)

= 4.78 W m⁻² s⁻¹

B )

Intensity of sound at 1.5 m distance = 60 /4 π d²

d = 1.5  m

Intensity of sound at 1 m distance = 60 /(4 π 1.5²)

= 2.12 W m⁻² s⁻¹

C )

Intensity of sound due to 4 speakers at 1.5 m distance = 4 x 60 /4 π d²

d = 1.5  m

= 4 x 60 /(4 π 1.5²)

= 8.48 W m⁻² s⁻¹

D )

Intensity of sound due to .06 W speaker must be 10⁻¹² W s ⁻² . Let the distance be d .

.06 /4 π d² = 10⁻¹²

d² = .06 /4 π 10⁻¹²

d = 6.9 x 10⁴ m .

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Feliz [49]
7. PE=0.5×700n/m×0.9m^2
0.9^2=0.81m
0.5×700×0.81= 283.5J

8. 2000=0.5×(x)×1.5m^2
1.5^2= 0.25
0.25×0.5=0.125
2000=0.125 (x)
2000/0.125=x
x=16000 n/m

9. 4000=0.5 (375 n/m)×(x)^2
0.5×187.5 (x)
4000/187.5=21.3333333333


6 0
3 years ago
Using Newton's third law of motion, explain what happens when you let an untied balloon go.
mash [69]

Answer:

balloon pushes you back

Explanation:

3rd Law: Every action has an equal and opposite reaction

So, when you let go of the balloon it's pushed forward so the balloon pushes you back

7 0
3 years ago
Read 2 more answers
Steam enters a counterflow heat exchanger operating at steady state at 0.07 MPa with a specific enthalpy of 2431.6 kJ/kg and exi
PIT_PIT [208]

Answer:

Explanation:

Given:

Steam Mass rate, ms = 1.5 kg/min

= 1.5 kg/min × 1 min/60 sec

= 0.025 kg/s

Air Mass rate, ma = 100 kg/min

= 100 kg/min × 1 min/60 sec

= 1.67 kg/s

A.

Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.

xf, quality = 0.9.

Tsat = 89.9°C

hf = 376.57 kJ/kg

hfg = 2283.38 kJ/kg

Using the equation for specific enthalpy,

hi = hf + (hfg × xf)

= 376.57 + (2283.38 × 0.9)

= 2431.552 kJ/kg

The specific enthalpy of the outlet, h2 = hf

= 376.57 kJ/kg

B.

Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy

= ms × (hi - h2)

= 0.025 × (2431.552 - 376.57)

= 0.025 × 2055.042

= 51.37455 kW

= 51.38 kW.

5 0
3 years ago
A block and tackle is used to lift an automobile engine that weighs 1800 N. The person exerts a force of 300 N to lift the engin
Sidana [21]

Answer:

1800/300 = 6ropes

Explanation:

The engine weighs 1800N and the person exerts a force of 300N, so for him to lift the engine and exerting a force of 300N all through we divide the weight of the engine by the force exerted to know how many ropes are used. Which makes it 6 thereby each rope uses 300N to lift the engine.

8 0
3 years ago
If we increase the speed of rotation of wheel then the sound produced by cardboard touching the spokes of the wheel will become
Anarel [89]

it will become shrill

5 0
3 years ago
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