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2 years ago

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An ice sheet 5m thick covers a lake that is 20m deep. what is the temperature of the water at the bottom of the lake? explain yo

ur answer . coefficient of volume expansivity at 20 degree centigrade is 2.1x10^-4

2 answers:

oksano4ka [1.4K]2 years ago

8 0

Answer:

4°C

Explanation:

An ice sheet 5m thick covers a lake that is 20m deep. The temperature of the water at the bottom of the lake is 4°C because it is dense, making it easy to sink.

Coefficient of volume expansivity at 20 degree centigrade is 2.1x10^-4.

umka2103 [35]2 years ago

5 0

Answer:

4°C

Explanation:

Water is densest at 4°C. Since dense water sinks, the bottom of the lake will be 4°C.

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Water is stored in a municipal water tank at a mean height of 25 m. If a faucet of diameter 1.2 cm is opened in a house at groun

ddd [48]

Answer:

22.14 m/s

Explanation:

height, h = 25 m

diameter, = 1.2 cm

Let v be the velocity of water.

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 25}$

v = 22.14 m/s

Thus, the speed of water is 22.14 m/s .

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2 years ago

100-ft-long horizontal pipeline transporting benzene develops a leak 43 ft from the high-pressure end. The diameter of the leak

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Answer:

Explanation:

The mass flow rate of benzene from the leak in the pipeline containing benzene is:

$Q_m=AC_o\sqrt{2\rho g_cP_g}$

Here, $Q_m$ is the mass flow rate through the leak of the pipeline. $A$ is the area of the hole, $C_o$ is the discharge rate, $\rho$ is the fluid density, $g_c$ is the gravitational constant and $P_g$ is the constant gauge pressure within the process unit.

The diametre of the leak (d) is 0.1 in. Convert from in to ft.

$d=(0.1 in)(\frac{1ft}{12in})\\=8.33\times 10^{-3}ft$

Calculate the area (A) of the hole. The area of the hole is.

$A=\frac{\pi d^2}{4}$

Substitute 3.14 for $\pi$ and $8.33\times 10^{-3}ft$ for d and calculate A.

$A=\frac{\pi d^2}{4}\\\\\frac{(3.14)(8.33\times 10^{-3})^2}{4}\\\\5.45\times 10^{-5}ft^2$

The specific gravity of benzene is 0.8794. Specific gravity is the ratio of th density of a substance to the density of a reference substance.

Specific gravity of benzene = density of benzenee/denity of reference substance

Rewrite the expression in terms of density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

Take the reference substance as water. Density of water is $62.4\frac{Ib_m}{ft^3}$ . Calculate density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

$=(0.8794)(62.4\frac{Ib_m}{ft^3})\\\\54.9\frac{Ib_m}{ft^3}$

Calculate the pressure at the point of leak. The pressure is the average of the pressure of the high and low pressure end. Write the expression to calculate the average pressure.

Upstream x distance from upstream pressure end

$P_g$ =+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

Calculate the distance from the downstream pressure end. The distance from upstream pressure end is 43 ft. Total of the pipe is 100 ft.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

The distance from upstream pressure end is 43 ft. Total length of the pipe is 100 ft. Substitute the values in the equation.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

= 100ft - 43ft = 57 ft

Substitute 50 psig for upstream, 43 ft fr distance from the upstream pressure end, 40 psig for downstream pressure, 57 ft for distance from the downstream pressure end, and 100 ft for the total length of the horizontal pipeline and calculate $P_g$ .

Upstream x distance from upstream pressure end

$P_g$ =+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

$=\frac{(50psig\times 43ft)+(40psig \times 57ft)}{100ft}\\\\=44.3psig$

Convert the pressure from psig to $Ib_f/ft^2$

$P_g=(44.3psig)(\frac{1\frac{Ib_f}{ft^2}}{1psig})(144\frac{in^2}{ft^2})\\\\=6,379.2\frac{Ib_f}{ft^2}$

The leak is like a sharp orifice. Take the value of the discharge coefficient as 0.61.

Substitute $5.45\times 10^{-5}ft^2$ for A. 0.61 for $C_o$ , $54.9\frac{Ib_m}{ft^3}$ for $\rho$ , $32.17\frac{ft.Ib_m}{Ib_f.s^2}$ for $g_c$ , and $6,379.2\frac{Ib_f}{ft^2}$ for $P_g$ and calculate $Q_m$

$Q_m=AC_o\sqrt{2\rho g_cP_g}\\\\=(5.45\times 10^{-5}ft^2)(0.61)\sqrt{2(54.9\frac{Ib_m}{ft^3})(32.17\frac{ft.Ib_m}{Ib_f.s^2})(6,379.2\frac{Ib_f}{ft^2})}\\\\(3.3245\times 10^{-5}ft^2)\sqrt{22,533,031.21\frac{Ib^2_m}{ft^4.s^2}}\\\\=0.158\frac{Ib_m}{s}$

The mass flow rate of benzene through the leak in the pipeline is $0.158\frac{Ib_m}{s}$

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2 years ago

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In a situation in which the smarter armiture is rubbing on the field the most commin reason are defective bushings. So, the ststement the techician A ssys that the brushings shojld be replaced is correct. Techinican A is right.

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2 years ago

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Ryan and Becca are moving a folding table out of the sunlight. A cup of lemonade, with the message 0.44 kg is on the table. Becc

jarptica [38.1K]

Calculate the weight of the table through the equation,

W = mg

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The components of this weight can be calculated through the equation,

Wx = W(sin θ)

and Wy = W(cos θ)

x - component:
Wx = W(sin θ)
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2 years ago

Not too long ago houses were protected from excessive currents by fuses rather than circuit breakers. sometimes a fuse blew out

olganol [36]

Answer: Resistance = $8.21 \times 10^{-8} \Omega$

The approximate diameter of a penny is, <em>d</em> = 20 mm

thickness of penny is, <em>L = </em> 1.5× $10^{-3}$ mm

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= 3.14× $10^{-4}$ m²

The resistivity of copper is <em>ρ</em> = 1.72 x 10-8 Ωm.

Resistance,

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2 years ago