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2 years ago

8

How many squares titles (20cm x 20cm) are needed to coat the sides and base of a pool which is 10m long, 6 meter wide and 3m dee

p?

1 answer:

Zanzabum2 years ago

7 0

Answer:

3900 tiles

Explanation:

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A small family home in Tucson, Arizona has a rooftop area of 2667 square feet, and it is possible to capture rain falling on abo

Kazeer [188]

Answer:

volume = 53.747 m3 = 14198.138 gal

weight = 526652 N = 118396.08 lbf

Explanation:

We know that volume of water

$volume = A'\times H$

where A' = 61% of A

$= 0.61\times 2667 = 1626.87 sq ft$

$volume = 1626.87 \times (\frac{14}{12} ft)$

=1898.015 ft^3

$in\ m^3 = \frac{ 1898.015}{35.315} = 53.7457 m^3$

$in\ gallon = 1898.015 \times 7.481 = 14198.138 gallon$

$weight = \rho Vg$

$= 1000\times 53.74\times 9.8$

=526652 N

$In\ lbf = \frac{526652}{4.448} = 118396.08 lbf$

7 0

3 years ago

11. Technicians A and B are discussing

algol [13]

Answer:

C. Neither Technician A nor B

Explanation:

Just took the test

5 0

3 years ago

A strain gauge with a 5 mm gauge length gives a displacement reading of 1.25 um. Calculate the stress value given by this displa

KengaRu [80]

Answer:

stress = 50MPa

Explanation:

given data:

Length of strain guage is 5mm

displacement $\delta = 1.25 \mu m =\frac{1.25}{1000} = 0.00125 mm$

stress due to displacement in structural steel can be determined by using following relation

$E =\frac{stress}{strain}$

$stress = E \times strain$

where E is young's modulus of elasticity

E for steel is 200 GPa

$stress = 200\times 10^3 *\frac{1.25*10^{-3}}{5}$

stress = 50MPa

7 0

3 years ago

A hollow pipe is submerged in a stream of water so that the length of the pipe is parallel to the velocity of the water. If the

Arlecino [84]

Answer:

increases by a factor of 6.

Explanation:

Let us assume that the initial cross sectional area of the pipe is A m² while the initial velocity of the water is V m/s², hence the flow rate of the water is:

Initial flow rate = area * velocity = A * V = AV m³/s

The water speed doubles (2V m/s) and the cross-sectional area of the pipe triples (3A m²), hence the volume flow rate becomes:

Final flow rate = 2V * 3A = 6AV m³/s = 6 * initial flow rate

Hence, the volume flow rate of the water passing through it increases by a factor of 6.

8 0

3 years ago

Wet steam at 15 bar is throttled adiabatically in a steady-flow process to 2 bar. The resulting stream has a temperature of 130°

cricket20 [7]

Answer:

$\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}$

Explanation:

The adiabatic throttling process is modelled after the First Law of Thermodynamics:

$m\cdot (h_{in} - h_{out}) = 0$

$h_{in} = h_{out}$

Properties of water at inlet and outlet are obtained from steam tables:

State 1 - Inlet (Liquid-Vapor Mixture)

$P = 1500\,kPa$

$T = 198.29\,^{\textdegree}C$

$h = 2726.9\,\frac{kJ}{kg}$

$s = 6.3068\,\frac{kJ}{kg\cdot K}$

$x = 0.967$

State 2 - Outlet (Superheated Vapor)

$P = 200\,kPa$

$T = 130\,^{\textdegree}C$

$h = 2726.9\,\frac{kJ}{kg}$

$s = 7.1776\,\frac{kJ}{kg\cdot K}$

The change of entropy of the steam is derived of the Second Law of Thermodynamics:

$\Delta s = 7.1776\,\frac{kJ}{kg\cdot K} - 6.3068\, \frac{kJ}{kg\cdot K}$

$\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}$

6 0

3 years ago