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DedPeter [7]
3 years ago
5

. The inner and outer surfaces of a 4-m × 7-m brick wall of thickness 30 cm and thermal conductivity 0.69 W/m-K are maintained a

t temperatures of 20 °C and 5 °C, respectively. Determine the rate of heat transfer through the wall, in W.
Physics
1 answer:
Anni [7]3 years ago
4 0

Answer:

\frac{dQ}{dt} = 966 W

Explanation:

As we know that the rate of heat transfer due to temperature difference is given by the formula

\frac{dQ}{dt} = \frac{KA(\Delta T)}{L}

here we know that

K = 0.69 W/m-K

A = 4 m x 7 m

thickness = 30 cm

temperature difference is given as

\Delta T = 20 - 5 = 15 ^oC

now we have

\frac{dQ}{dt} = \frac{(0.69W/m-K)(28 m^2)(15)}{0.30}

\frac{dQ}{dt} = 966 W

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Answer:

true

Explanation:

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You are standing 2.5m directly in front of one of the two loudspeakers. They are 3.0m apart and both are playing a 686Hz tone in
ahrayia [7]

Answer:

distance from speaker is 17.87 m

Explanation:

given data

distance from loudspeaker = 2.5 m

distance between loudspeaker = 3.0 m

room temperature = 20c

wavelength f  = 686Hz

to find out

what distances from the speaker

solution

we know sound velocity c = 331.5  + 0.6 × 20c = 343.5

so wavelength of sound  λ = c / f  

wavelength = 343.5 /  686 = 0.5 m

when the difference in distance of speaker destructive interference will be

d = λ/2 × (2n-1)

for n = 1, 2 3 4 ..

d = 0.5/2 × (2n-1)

d = 0.250 , 0.75 , 1.25 , 1.750............   for n = 1, 2 3 .............

so

for d = 0.250

side of triangle by hypotenuse of triangle are

\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x1) = 0.250

0.5 x1 = 7.6875

x1 = 15.375 m

for d = 0.75

side of triangle by hypotenuse of triangle are

\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x2) = 0.75

1.5 x2 = 4.6875

x2 = 3.125 m

for d = 1.250

side of triangle by hypotenuse of triangle are

\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x3) = 1.250

2.5 x2 = 1.1875

x3 = 0.475 m

for d = 1.750

x4 will be negative so we stop here

so the distance from speaker here is given below

distance = 2.5 + x

here x = 0.475 , 3.125 and 15.375 so

distance 1 = 2.5 + 0.475  = 2.975 m

distance 2 = 2.5 + 3.125  = 5.625 m

distance 3 = 2.5 + 15.375 = 17.875 m

final distance from speaker is 17.87 m

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