I need the answer please

The quantity of bricks required increases with the surface area of the wall, but the thickness of a masonry wall does not affect

Yakvenalex [24]

Answer:

false

Explanation:

5 0

3 years ago

Read 2 more answers

In a cellular phone system, a mobile phone must be paged to receive a phone call. However, paging attempts don’t always succeed

Alina [70]

Answer:

The correct response will be "0.992". The further explanation to the following question is given below.

Explanation:

The probability that paging would be beneficial becomes 0.8

Effective paging at the very first attempted is 0.8

On the second attempt the success probability will be:

⇒ $0.2\times 0.8$

⇒ $0.16$

On the third attempt the success probability will be:

⇒ $0.2\times 0.2\times 0.8$

⇒ $0.032$

So that the success probability will be:

⇒ $0.8 + 0.16 + 0.032$

⇒ $0.992$

7 0

3 years ago

Water flows at a rate of 10 gallons per minute in a new horizontal 0.75?in. diameter galvanized iron pipe. Determine the pressur

ruslelena [56]

Answer:

$\frac{\delta p }{l} = 30.4 lb/ft^3$

Explanation:

Given data:

flow rate = 10 gallon per minute = 0.0223 ft^3/sec

diameter = 0.75 inch

we know discharge is given as

Q = VA

solve for velocity V = \frac{Q}{A}[/tex]

$V = \frac{0.223}{\frac{\pi}{4} \frac{0.75}{12}}$

V = 7.27 ft/sec

we know that Reynold number

$Re = \frac{VD}{\nu}$

$Re = \frac{7.27 \times \frac{0.75}{12}}{1.21\times 10^{-5}}$

$Re = 3.76 \times 10^4$

calculate the $\frac{\epsilon }{D}$ ratio to determine the fanning friction f

$\frac{\epsilon }{D} = \frac{0.0005}{\frac{0.75}{12}} = 0.008$

from moody diagram f value corresonding to Re and $\frac{\epsilon }{D}$ is 0.037

for horizontal pipe

$\delta p = \frac{f l \rho v^2}{2D}$

$\frac{\delta p }{l} = \frac{1 \times 0.037 \times 1.94 \times 7.27}{\frac{0.75}{12}}$

where 1.94 slug/ft^3is density of water

$\frac{\delta p }{l} = 30.4 lb/ft^3$

3 0

3 years ago

How would I find the Voltage across the open circuit

Nimfa-mama [501]

Answer:

Vab = 80V

Explanation:

The only current flowing in the circuit is supplied by the 100 V source. Its only load is the 40+60 ohm series circuit attached, so the current in that loop is (100V)/(40+60Ω) = 1A. That means V1 = (1A)(60Ω) = 60V.

Vab will be the sum of voltages around the right-side "loop" between terminals 'a' and 'b'. It is (working clockwise from terminal 'b') ...

Vab = -10V +60V +(0A×10Ω) +30V

Vab = 80V

3 0

2 years ago

A piston–cylinder device contains a mixture of 0.5 kg of H2 and 1.2 kg of N2 at 100 kPa and 300 K. Heat is now transferred to th

Taya2010 [7]

Answer:

(a) The heat transferred is 2552.64 kJ

(b) The entropy change of the mixture is 1066.0279 J/K

Explanation:

Here we have

Molar mass of H₂ = 2.01588 g/mol

Molar mass of N₂ = 28.0134 g/mol

Number of moles of H₂ = 500/2.01588 = 248 moles

Number of moles of N₂ = 1200/28.0134 = 42.8 moles

P·V = n·R·T

V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³

Since the volume is doubled then

V₂ = 2 × 7.25 = 14.51 m³

At constant pressure, the temperature is doubled, therefore

T₂ = 600 K

If we assume constant specific heat at the average temperature, we have

Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂

cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K

cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K

Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 = 2552.64 kJ

b) $\Delta S = - R(n_A \times lnx_A + n_B \times ln x_B)$

Where:

$x_A$ and $x_B$ are the mole fractions of Hydrogen and nitrogen respectively.

Therefore, $x_A$ = 248 /(248 + 42.8) = 0.83

$x_B$ = 42.8/(248 + 42.8) = 0.1472

∴ $\Delta S = - 8.3145(248 \times ln0.83 + 42.8 \times ln 0.1472)$ = 1066.0279 J/K

5 0

3 years ago