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Grace [21]
3 years ago
6

Thermodynamics fill in the blanks The swimming pool at the local YMCA holds roughly 749511.5 L (749511.5 kg) of water and is kep

t at a temperature of 80.6 °F year round using a natural gas heater. If you were to completely drain the pool and refill the pool with 50°F water, (blank) GJ (giga-Joules) of energy are required to to heat the water back to 80.6 °F. Note: The specific heat capacity of water is 4182 J/kg ⋅°C. The cost of natural gas per GJ is $2.844. It costs $ (blank) to heat the pool (to the nearest dollar).
Engineering
1 answer:
Talja [164]3 years ago
7 0

Answer:

95.914\ \text{GJ}

\$272.78

Explanation:

m = Mass of water = 749511.5 kg

c = Specific heat of water = 4182 J/kg ⋅°C

\Delta T = Change in temperature = 80.6-50=30.6^{\circ}\text{F}

Cost of 1 GJ of energy = $2.844

Heat required is given by

Q=mc\Delta T\\\Rightarrow Q=749511.5\times 4182\times 30.6\\\Rightarrow Q=95.914\times 10^9\ \text{J}=95.914\ \text{GJ}

Amount of heat required to heat the water is 95.914\ \text{GJ}.

Cost of heating the water is

95.914\times 2.844=\$272.78

Cost of heating the water to the required temperature is \$272.78.

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Instead of running blood through a single straight vessel for a distance of 2 mm, one mammalian species uses an array of 100 tin
Marina CMI [18]

Solution:

Given that :

Volume flow is, $Q_1 = 1000 \ mm^3/s$

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Therefore, the equation of a single straight vessel is given by

$F_{f_1}=\frac{8flQ_1^2}{\pi^2gd_1^5}$    ......................(i)

So there are 100 similar parallel pipes of the same cross section. Therefore, the equation for the area is

$\frac{\pi d_1^2}{4}=1000 \times\frac{\pi d_2^2}{4} $

or $d_1=10 \ d_2$

Now for parallel pipes

$H_{f_2}= (H_{f_2})_1= (H_{f_2})_2= .... = = (H_{f_2})_{10}=\frac{8flQ_2^2}{\pi^2 gd_2^5}$  ...........(ii)

Solving the equations (i) and (ii),

$\frac{H_{f_1}}{H_{f_2}}=\frac{\frac{8flQ_1^2}{\pi^2 gd_1^5}}{\frac{8flQ_2^2}{\pi^2 gd_2^5}}$

       $=\frac{Q_1^2}{Q_2^2}\times \frac{d_2^5}{d_1^5}$

       $=\frac{(1000)^2}{(10)^2}\times \frac{d_2^5}{(10d_2)^5}$

       $=\frac{10^6}{10^7}$

Therefore,

$\frac{H_{f_1}}{H_{f_2}}=\frac{1}{10}$

or $H_{f_2}=10 \ H_{f_1}$

Thus the answer is option A). 10

7 0
3 years ago
If the bending moment (M) is 4,176 ft-lb and the beam is an 1 beam, calculate the bending stress (psi) developed at a point with
SpyIntel [72]

Answer:

Bending stress at point 3.96 is \sigma_b = 1.37 psi

Explanation:

Given data:

Bending Moment M is 4.176 ft-lb = 50.12 in- lb

moment of inertia I = 144 inc^4

y = 3.96 in

\sigma_b = \frac{M}{I} \times y

putting all value to get bending stress

\sigma_b = \frac{50.112}{144} \times 3.96  

\sigma_b =  1.37 psi

Bending stress at point 3.96 is \sigma_b = 1.37 psi

3 0
4 years ago
A data record of all of someone's online activity is called a
aev [14]

Answer:

The answer is computer cookies.

Explanation:

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4 0
3 years ago
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Find the derivative of y = sin(ln(5x2 − 2x))
pickupchik [31]

Answer:

y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)

Explanation:

Let y = \sin[\ln(5\cdot x^{2}-2\cdot x)] and we proceed to find the derivative by the following steps:

1) y = \sin[\ln(5\cdot x^{2}-2\cdot x)] Given

2) y = \sin [\ln[x\cdot (5\cdot x - 2)]] Distributive property

3) y = \sin[\ln x + \ln (5\cdot x - 2 )] \ln (a\cdot b) = \ln a + \ln b

4) y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)  \frac{d}{dx} (\sin x) = \cos x/\frac{d}{dx}(\ln x) = \frac{1}{x}/\frac{d}{dx}(c\cdot x^{n}) = n\cdot c\cdot x^{n-1}/Rule of chain/Result

3 0
3 years ago
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Oxana [17]

Answer and Explanation:

A warehouse can be defined as a place:

is a spot or building, where the articles are kept, put away or in some cases prepared.  

Depending on the sort of use, the checkpoints may fluctuate.  

There can be numerous sorts, for example, pressing, railroad wagon, cool stockpiling, and so on.  

Suppositions are caused in regards to each shape and a few checkpoints for an item to can be summed up.  

  • The point of entry, where it ought to be through into the product house.  
  • Security checkpoint, where the item belonging is to be protected.  
  • Palletized segment, where the item is to be put in a spot and it determines the location of that item in the house.  
  • Quality or bundling segment, where the item must be tried and improved.  
  • Leave segment, through which it must be taken out for further procedures.  

7 0
4 years ago
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