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4 years ago

12

How are density, specific weight, and specific gravity related? (in words)

1 answer:

bezimeni [28]4 years ago

8 0

Answer:

Specific gravity is an expression of density in relation to the density of a standard or reference (usually water). Also, density is expressed in units (weight relative to size) while specific gravity is a pure number or dimensionless.

Explanation:

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Lawn maintenance is an alternative energy source<br> -true<br> -false

Reika [66]

Answer:

false

Explanation:

4 0

3 years ago

A car is moving at 68 miles per hour. The kinetic energy of that car is 5 × 10 5 J.How much energy does the same car have when i

Blababa [14]

Answer:

The car has an energy of 1.017 × 10⁵ J when it moves at 97 miles per hour

Explanation:

Given the data in the question;

Initial velocity v₁ = 68 miles per hour = 30.398 meter per seconds

let mass of the car be m

kinetic energy of that car is 5 × 10⁵ J

so

E₁ = $\frac{1}{2}$ mv²

we substitute

5 × 10⁵ = $\frac{1}{2}$ × m × ( 30.398 )²

5 × 10⁵ = $\frac{1}{2}$ × m × ( 30.398 )²

5 × 10⁵ = m × 462.019

m = 5 × 10⁵ / 462.019

m = 1082.2065 kg

Now, Also given that; v₂ = 97 miles per hour = 43.362 meter per seconds

E₂ = $\frac{1}{2}$ mv₂²

we substitute

E₂ = $\frac{1}{2}$ × 1082.2065 × ( 43.362 )²

E₂ = $\frac{1}{2}$ × 1082.2065 × 1880.263

E₂ = 1.017 × 10⁵ J

Therefore, The car has an energy of 1.017 × 10⁵ J when it moves at 97 miles per hour

6 0

3 years ago

25°C is flowing in a covered irrigation ditch below ground. Every 100 ft, there is a vent line with a 1.0 in. inside diameter an

Gnesinka [82]

The total evaporation loss of water is 87.873 × $10^{-5}$ lbm /day.

<u>Explanation:</u>

Assume A is the water and B is air.

A is diffusing to non diffusing B.

$N_{A}=\frac{D_{A B} P}{R T Z P_{B/ M}}\left(P_{A_{1}} \ - P_{B_{2}}\right)$

By the table 6.2 - 1 at 25°C, the diffusivity of air and water system is $0.260 \times 10^{-4} \mathrm{m}^{2} / \mathrm{s}$ .

Total pressure P = 1 atm = 101.325 KPa

$P_{A_{1} }$ = 23.76 mm Hg

$P_{A_{1} }$ = $\frac{23.76}{760}$

$P_{A_{1} }$ = 0.03126 atm

$P_{A_{1} }$ = 3.167 K Pa

When air surrounded is dry air, then $PA_{2}$ = 0 mm Hg

R = 8.314 $\frac{K P a \cdot m^{3}}{mol \cdot k}$

$P_{B/M}$ = $\frac{P_{B_{1}}-P_{B_{2}}}{\ln \left(P_{B_{1}} / P_{B_{2}}\right)}$

$P_{B_{1}}=P_{T}-P_{A1_{}}$

= 101.325 - 3.167

$P_{B_{1} }$ = 98.158 K Pa

$P_{B_{2}}=P_{T}-P_{A_{2}}$

$P_{B_{2} }$ = 101.325 - 0

$P_{B_{2} }$ = 101.325 K Pa

$P_{B / M}=\frac{98.158-101.325}{\ln (98.158 / 101 \cdot 325)}$

$P_{B/M}$ = 99.733 K Pa

Z = 1 ft = 0.3048 m

T = 298 K

$N_{A}$ = $\frac{(0.206*10^{-4})(101.325)(3.167 - 0) }{(8.314) ( 298 )( 0.3048 )( 99.733 ) }$

$N_{A}$ = 0.11077 × $10^{-6}$ mol/ $m^{2}$ .s

$N_{A}$ = 0.11077 × $10^{-6}$ × 18 × (60×60×24)

$N_{A}$ = 0.1723 $lb_{m}$ / $m^{2}$ .day

$\tilde{N}_{A}=N_{A} \times Area$

Area of individual pipe is

$A=\frac{\pi}{4}(0.0254)^{2}$

A = 0.00051 $m^{2}$

$\bar{N}_{\boldsymbol{A}}$ = 0.1723 × 0.00051

$\bar{N}_{\boldsymbol{A}}$ = 0.000087873 lbm/day

In 1000 ft length of ditch,there will be a 10 pipes. The amount of evaporation water is

= 10 × 0.000087873 = 0.00087873 lbm /day

The total evaporation loss of water is 87.873 × $10^{-5}$ lbm /day.

5 0

3 years ago

Description: An apartment building has a pumped ring main that circulates hot water at 75 C through insulated 40 mm pipe with he

Ratling [72]

Answer:

hfhhfhff

Explanation:

7 0

3 years ago

A piston-cylinder device contains 1.329 kg of nitrogen gas at 120 kPa and 27 degree C. The gas is now compressed slowly in a pol

Shtirlitz [24]

Answer:-0.4199 J/k

Explanation:

Given data

mass of nitrogen(m)=1.329 Kg

Initial pressure $\left ( P_1\right )$ =120KPa

Initial temperature $\left ( T_1\right )=27\degree \approx$ 300k

Final volume is half of initial

R=particular gas constant

Therefore initial volume of gas is given by

PV=mRT

V=0.986\times 10^{-3}

Using $PV^{1.49}$ =constant

$P_{1}V^{1.49}$ = $P_2\left (\frac{V}{2}\right )$

$P_2$ =337.066KPa

$V_2$ = $0.493\times 10^{-3} m^{3}$

and entropy is given by

$\Delta s$ = $C_v \ln \left (\frac{P_2}{P_1}\right )$ + $C_p \ln \left (\frac{V_2}{V_1}\right )$

Where, $C_v$ = $\frac{R}{\gamma-1}$ =0.6059

$C_p$ = $\frac{\gamma R}{\gamma -1}$ =0.9027

Substituting values we get

$\Delta s$ = $0.6059\times\ln \left (\frac{337.066}{120}\right )$ + $0.9027 \ln \left (\frac{1}{2}\right )$

$\Delta s$ =-0.4199 J/k

4 0

4 years ago