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3 years ago

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A small omnidirectional stereo speaker produces waves in all directions that have an intensity of 8.00 at a distance of 4.00 fro

m the speaker.
At what rate does this speaker produce energy?

What is the intensity of this sound 9.50 from the speaker?

What is the total amount of energy received each second by the walls (including windows and doors) of the room in which this speaker is located?

1 answer:

slavikrds [6]3 years ago

5 0

Answer:

A. We have that radius r = 4.00m intensity I = 8.00 W/m^

total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W

b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2

c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J

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A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of 5.64 10-2 kg/s. the density of the gasoline is 735

Irina18 [472]

Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
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Radius of fuel line, r = 3.43 x 10⁻³ m

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A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²

Let v = speed of pumping the gasoline, m/s
Then the mass flow rate is
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s

The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
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3 years ago

Read 2 more answers

1.- Un barco recorre la distancia que separa Gran Canaria de Tenerife (90 km) en 6 horas. ¿Cuál es

bezimeni [28]

Answer:

The speed is 15 km/h or 4.16 m/s.

Explanation:

A boat travels the distance that separates Gran Canaria from Tenerife (90 km) in 6 hours. Which the speed of the boat in km / h? And in m / s?

Given that,

Distance, d = 90 km = 90000 m

Time, t = 6 hours = 21600 s

Speed = distance/time

$v=\dfrac{90\ km}{6\ h}\\\\=15\ km/h$

or

$v=\dfrac{90000\ m}{21600\ s}\\\\=4.16\ m/s$

So, the required speed is 15 km/h or 4.16 m/s.

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2 years ago

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2 years ago

An inventor claims to have developed a food freezer that, in steady-state conditions, requires a power input of 0.25 kW to extra

WARRIOR [948]

Answer:

The inventors claim is not real

a) No the the freezer cannot operate in such conditions

Explanation:

From the question we are told that

The power input is $P_i = 0.25 kW = 0.25 *10^{3} \ W$

The rate of heat transfer $J = 3050 J/s$

The temperature of the freezer content is $T = 270 \ K$

The ambient temperature is $T_a = 293 \ K$

Generally the coefficient of performance of a refrigerator at idea conditions is mathematically represented as

$COP = \frac{T }{Ta - T}$

substituting values

$COP = \frac{270 }{293 - 270}$

$COP =11.7$

Generally the coefficient of performance of a refrigerator at real conditions is mathematically represented as

$COP = \frac{J}{P_i}$

substituting values

$COP = \frac{3050}{0.25 *10^{3}}$

$COP = 12.2$

Now given that the COP of an ideal refrigerator is less that that of a real refrigerator then the claims of the inventor is rejected

This is because the there are loss in the real refrigerator cycle that are suppose to reduce the COP compared to an ideal refrigerator cycle where there no loss that will reduce the COP

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2 years ago

A 2.4-m high 200-m2 house is maintained at 22Â°C by an air-conditioning system whose COP, is 3.2. It is estimated that the kitch

kotegsom [21]

Answer:

The amount that is "vented" out by "the fans" is <u>$0.50</u> for 10 hours.

Option: a

<u>Explanation</u>:

"Energy discharged by air in every hour" can be determined by,

$\mathrm{Q}=\mathrm{m}_{\mathrm{air}} \mathrm{C}_{\mathrm{p}} \Delta \mathrm{T}$

Q = heat energy (Joules, J)

m = mass of a substance (kg)

c = specific heat (units J/kg∙K)

$\mathrm{m}_{\mathrm{air}}=\rho \mathrm{v}$

$\text { Density of air } \rho=1.20 \mathrm{kg} / \mathrm{m}^{3}$

$\text { Density of air } \rho=1.20 \times 200 \times 2.4$

$\text { Density of air } \rho=576 \mathrm{kg}$

∆T = 10 hours

$\text { Specific Heat Capacities of Air. The nominal values used for air at } 300 \mathrm{K} \text { are } \mathrm{C_P}=1.00 \mathrm{kJ} / \mathrm{kg} . \mathrm{K}$

Q = 576 × 1.00 × 10

Q = 5760 kJ/hours

W = 1.6 kwh

We know that, “Coefficient of performance” (COP)

$\mathrm{Cop}=\frac{Q}{w}$

$\mathrm{W}=\frac{Q}{\mathrm{cop}}$

Given that, COP = 3.2

$\mathrm{W}=\frac{1.6}{3.2}$

W = 0.5 kwh

The unit cost of electricity is $0.10/kWh

The unit cost of electricity is $0.10/kWh

Unit electricity cost for 10 hours = 0.5 × 10 × 0.1$

Unit electricity cost for 10 hours = $0.5

The amount that is "vented out" by "the fans" is $0.50 for 10 hours.

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3 years ago