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Burka [1]
3 years ago
10

Write an algorithm and draw a flowchart to convert the length in feet to centimeter

Engineering
1 answer:
Simora [160]3 years ago
3 0

Answer:

blah blah blah sh ut up read learn

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A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp
cricket20 [7]

Answer:

Given that;

Jello there, see explanstion for step by step solving.

A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.

Explanation:

A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.

See attachment for more clearity

6 0
3 years ago
A three-point bending test is performed on a glass specimen having a rectangular cross section of height d 5 mm (0.2 in.) and wi
Anon25 [30]

Answer:

The flexural strength of a specimen is = 78.3 M pa

Explanation:

Given data

Height = depth = 5 mm

Width = 10 mm

Length L = 45 mm

Load = 290 N

The flexural strength of a specimen is given by

\sigma = \frac{3 F L}{2 bd^{2} }

\sigma = \frac{3(290)(45)}{2 (10)(5)^{2} }

\sigma = 78.3 M pa

Therefore the flexural strength of a specimen is = 78.3 M pa

4 0
3 years ago
A smooth sphere with a diameter of 6 inches and a density of 493 lbm/ft^3 falls at terminal speed through sea water (S.G.=1.0027
Pachacha [2.7K]

Given:

diameter of sphere, d = 6 inches

radius of sphere, r = \frac{d}{2} = 3 inches

density,  \rho} = 493 lbm/ ft^{3}

S.G = 1.0027

g = 9.8 m/ m^{2} = 386.22 inch/ s^{2}

Solution:

Using the formula for terminal velocity,

v_{T} = \sqrt{\frac{2V\rho  g}{A \rho C_{d}}}              (1)

(Since, m = V\times \rho)

where,

V = volume of sphere

C_{d} = drag coefficient

Now,

Surface area of sphere, A = 4\pi r^{2}

Volume of sphere, V = \frac{4}{3} \pi r^{3}

Using the above formulae in eqn (1):

v_{T} = \sqrt{\frac{2\times \frac{4}{3} \pir^{3}\rho  g}{4\pi r^{2} \rho C_{d}}}

v_{T} = \sqrt{\frac{2gr}{3C_{d}}}  

v_{T} = \sqrt{\frac{2\times 386.22\times 3}{3C_{d}}}

Therefore, terminal velcity is given by:

v_{T} = \frac{27.79}{\sqrt{C_d}} inch/sec

3 0
3 years ago
Refrigerant 134a enters an air conditioner compressor at 4 bar, 20 C, and is compressed at steady state to 12 bar, 80 C. The vol
sleet_krkn [62]

Answer:

Q=15.7Kw

Explanation:

From the question we are told that:

Initial Pressure P_1=4bar

Initial Temperature T_1=20 C

Final Pressure  P_2=12 bar

Final Temperature T_2=80C

Work Output W= 60 kJ/kg

Generally Specific Energy from table is

At initial state

 P_1=4bar \& T_1=20 C

 E_1=262.96KJ/Kg

With

Specific Volume V'=0.05397m^3/kg

At Final state

 P_2=12 bar \& P_2=80C

 E_1=310.24KJ/Kg

Generally the equation for The Process is mathematically given by

 m_1E_1+w=m_2E_2+Q

Assuming Mass to be Equal

 m_1=m_1

Where

 m=\frac{V}{V'}

 m=frac{0.06666}{V'=0.05397m^3/kg}

 m=1.24

Therefore

 1.24*262.96+60)=1.24*310.24+Q

 Q=15.7Kw

4 0
3 years ago
Technician A says that after replacing a power steering hose, the system should be flushed, refilled, and bled. Technician B say
dedylja [7]
The. Answer will be D
3 0
3 years ago
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