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3 years ago

14

What is the speaker’s power output if the sound intensity level is 102 dBdB at a distance of 25 mm ? Express your answer to two

significant figures and include the appropriate units.

1 answer:

lesya [120]3 years ago

7 0

Answer:

Power = 124.50 W

Explanation:

Given that:

The Sound intensity of a speaker output is 102 dB

and the distance r = 25 m

For the intensity of sound,

$\beta (dB)= 10 \ log_{10 } (\dfrac{I}{I_o})$

where;

the threshold of hearing $I_o = 10^{-12} (W/m^2)$

$\dfrac{102 }{10}= log_{10}( \dfrac{I}{10^{-12}})$

$10^{10.2} = \dfrac{I}{10^{-12}}$

$I = 10^{10.2} \times 10^{-12}$

I = 0.01585 W/m²

If we recall, we know remember that ;

Power = Intensity × A rea

Power = 0.01585 W/m² × 4 × 3.142 × (25 m)²

Power = 124.50 W

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natita [175]

Answer:

Final temperature, T2 = 314.9 Kelvin

Explanation:

Given the following data:

Mass = 0.9kg

Initial temperature, T1 = 10°C to Kelvin = 10 + 273 = 283K

Quantity of heat = 120,000 J

Specific heat capacity = 4182 j/kgK

To find the final temperature;

Heat capacity is given by the formula;

$Q = mcdt$

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

Making dt the subject of formula, we have;

$dt = \frac {Q}{mc}$

Substituting into the equation, we have;

$dt = \frac {120000}{0.9*4182}$

$dt = \frac {120000}{3763.8}$

dt = 31.9K

Now, the final temperature T2 is;

But, dt = T2 - T1

T2 = dt + T1

T2 = 31.9 + 283

T2 = 314.9 Kelvin

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3 years ago

A heat engine has a maximum possible efficiency of 0.780. If it operates between a deep lake with a constant temperature of-24.8

Volgvan

Answer : The temperature of the hot reservoir (in Kelvins) is 1128.18 K

Explanation :

Efficiency of carnot heat engine : It is the ratio of work done by the system to the system to the amount of heat transferred to the system at the higher temperature.

Formula used for efficiency of the heat engine.

$\eta =1-\frac{T_c}{T_h}$

where,

$\eta$ = efficiency = 0.780

$T_h$ = Temperature of hot reservoir = ?

$T_c$ = Temperature of cold reservoir = $-24.8^oC=273+(-24.8)=248.2K$

Now put all the given values in the above expression, we get:

$\eta =1-\frac{T_c}{T_h}$

$0.780=1-\frac{248.2K}{T_h}$

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3 years ago

. Reem took a wire of length 10 cm. Her friend Nain took a wire of 5 cm of the same material and thickness both of them connecte

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Given :

Reem took a wire of length 10 cm. Her friend Nain took a wire of 5 cm of the same material and thickness both of them connected with wires as shown in the circuit given in figure. The current flowing in both the circuits is the same.

To Find :

Will the heat produced in both the cases be equal.

Solution :

Heat released is given by :

H = i²Rt

Here, R is resistance and is given by :

$R = \dfrac{\rho L}{A}$

So,

$H = i^2\times \dfrac{\rho L}{A} t\\\\H = \dfrac{i^2\rho Lt}{A}$

Now, in the question every thing is constant except for the length of the wire and from above equation heat is directly proportional to the length of the wire.

So, heat produced by Reem's wire is more than Nain one.

Hence, this is the required solution.

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3 years ago

An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and

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Answer:

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Explanation:

Given that:

Total Length for aluminum bar = 600 mm

Diameter for aluminum bar = 40 mm

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Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

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Calculate the total contraction on the bar = ???

The relation used in calculating the contraction on the bar is:

$\delta L = \dfrac{P *L }{A*E}$

The relation used in calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = $\dfrac{P *L_1 }{A_1*E} + \dfrac{P *L_2 }{A_2 *E}$

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

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A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

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Total contraction on the Bar = $\dfrac{180 *10^3 \N }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]$

Total contraction on the Bar = 2.117( 0.398 + 0.182)

Total contraction on the Bar = 2.117*(0.58)

Total contraction on the Bar = 1.22786 mm

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3 years ago

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Naddika [18.5K]

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3 years ago