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Marat540 [252]
3 years ago
10

A heat pump receives heat from a lake that has an average winter time temperature of 6o C and supplies heat into a house having

an average temperature of 25o C. (a) If the house loses heat to the atmosphere at the rate of 60,000 kJ/h, determine the minimum power supplied to the heat pump (in kW) that can maintain the interior temperature of the house at 25o C. (b) Suppose the heat pump absorbs energy from the ground (10o C) instead of the lake, Calculate the minimum power (in kW) required for the heat pump.
Physics
1 answer:
Natalija [7]3 years ago
4 0

Answer:

Explanation:

60000 kJ / h = 60000 x 1000 / (60 x 60 )

= 16667 J /s

a)

\frac{Q_H}{W} =\frac{T_H}{T_H-T_C}

where Q_H and W are heat supplied and heat given from  outside source .

Here Q_H = 16667 J/s

\frac{T_H}{T_H-T_C}=\frac{298}{298-279}

= 15.684

Q_H = 16667 J/s

16667 / W = 15.684

W= minimum power supplied  = 1062.7 W.

b )

If  T_C=283K

\frac{T_H}{T_H-T_C}=\frac{298}{298-283}

=19.87

\frac{Q_H}{W} = 19.87

W=\frac{Q_H}{19.87}

W=\frac{16667}{19.87}

= 838.8 W .

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Fofino [41]

Answer:

(a) Wavelength is 0.436 m

(b) Length is 0.872 m

(c) 11.518 m/s

Solution:

As per the question:

The eqn of the displacement is given by:

y(x, t) = (1.22 cm)sin[14.4 m^{- 1}x]cos[(166\ rad/s)t]          (1)

n = 4

Now,

We know the standard eqn is given by:

y = AsinKxcos\omega t           (2)

Now, on comparing eqn (1) and (2):

A = 1.22 cm

K = 14.4 m^{- 1}

\omega = 166\ rad/s

where

A = Amplitude

K = Propagation constant

\omega = angular velocity

Now, to calculate the string's wavelength,

(a) K = \frac{2\pi}{\lambda}

where

K = propagation vector

\lambda = \frac{2\pi}{K}

\lambda = \frac{2\pi}{14.4} = 0.436\ m

(b) The length of the string is given by:

l = \frac{n\lambda}{2}

l = \frac{4\times 0.436}{2} = 0.872\ m

(c)  Now, we first find the frequency of the wave:

\omega = 2\pi f

f = \frac{\omega}{2\pi}

f = \frac{2\pi}{166} = 26.42\ Hz

Now,

Speed of the wave is given by:

v = f\lambda

v = 26.419\times 0.436 = 11.518\ m/s

4 0
3 years ago
At a given instant the bottom A of the ladder has an acceleration aA = 4 f t/s2 and velocity vA = 6 f t/s, both acting to the le
Nana76 [90]

Answer:

Acceleration=24.9ft^2/s^2

Angular acceleration=1.47rads/s

Explanation:

Note before the ladder is inclined at 30° to the horizontal with a length of 16ft

Hence angular velocity = 6/8=0.75rad/s

acceleration Ab=Aa +(Ab/a)+(Ab/a)t

4+0.75^2*16+a*16

0=0.75^2*16cos30°-a*16sin30°---1

Ab=0+0.75^2sin30°+a*16cos30°----2

Solving equation 1

(0.75^2*16cos30/16sin30)=angular acceleration=a=1.47rad/s

Also from equation 2

Ab=0.75^2*16sin30+1.47*16cos30=24.9ft^2/s^2

6 0
3 years ago
Hello please help i’ll give brainliest
Zarrin [17]

Answer:

C because it is in earth's mantle

3 0
3 years ago
What is the wavelength in nm of a light whose first order bright band forms a diffraction angle of 30 degrees, and the diffracti
Artist 52 [7]

Answer:

The wavelength is 3500 nm.

Explanation:

d= \frac{1}{700 lines per mm} = 0.007mm = 7000 nm

n= 1

θ= 30°

λ= unknown

Solution:

d sinθ = nλ

λ = \frac{7000 nm sin 30}{1}

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8 0
3 years ago
The half life of Po-218 is three minutes. How much of a 2.0 gram sample remains after 15 minutes? Suppose you wanted to buy some
Greeley [361]

After 15 minutes :

M = M₀ (1/2)^(t/T)

M = 2 (1/2)^(15/3)

M = 0.0625 gr

In order to get 0.1 gr :

M = M₀ (1/2)^(t/T)

0.1 = M₀ (1/2)^(30/3)

0.1 = Mo / 1024

Mo = 102.4 gr

5 0
2 years ago
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