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choli [55]
3 years ago
5

What could happen in the aviation

Engineering
1 answer:
Alina [70]3 years ago
7 0

Answer:

Yes

Explanation:

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Shortly after the introduction of a new​ coin, newspapers published articles claiming the coin is biased. The stories were based
garik1379 [7]

Answer:

(a) 0.12924

(b) Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.

Explanation:

(a)

n=200 for fair coin getting head, p= 0.5

Expectation = np =200*0.5=100

Variance = np(1 - p) = 100(1-0.5)=100*0.5=50

Standard deviation, s = \sqrt {variance}=\sqrt {50}= 7.071068

Z value for 108, z =\frac {108-100}{7.071068}= 1.131371

P( x ≥108) = P( z >1.13)= 0.12924

(b)

Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.

3 0
3 years ago
In a pipe of diameter 350mm and length 75mm, water is flowing at a velocity of 2.8m/s. If the Kinematic viscosity of water is 0.
kodGreya [7K]

Explanation:

the answer is in picture

5 0
3 years ago
A lake is fed by a polluted stream and a sewage outfall. The stream and sewage wastes have a decay rate coefficient (k) of 0.5/d
joja [24]

Solution :

Given :

k = 0.5 per day

$C_s = 10 \ mg/L \ ; \ \ Q_s= 40 \ m^3/s$

$C_{sw} = 100 \ ppm \ ; \ \ Q_{sw}= 0.5 \ m^3/s$

Volume, V $= 200 \ m^3$

Now, input rate = output rate + KCV ------------- (1)

Input rate  $= Q_s C_s + Q_{sw}C_{sw}$

                $=(40 \times 10) + (0.5\times 100)$

                $= 2 \times 10^5 \ mg/s$

The output rate $= Q_m C_{m}$

                          = ( 40 + 0.5 ) x C x 1000

                          $=40.5 \times 10^3 \ C \ mg/s$

Decay rate = KCV

∴$KCV =\frac{0.5/d \times C \  \times 200 \times 1000}{24 \times 3600}$

            = 1.16 C mg/s

Substituting all values in (1)

$2 \times 10^5 = 40.5 \times 10^3 \ C+ 1.16 C$

C = 4.93 mg/L

4 0
3 years ago
Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2 /s is being discharged by a 5 mm diameter, 40 m long horiz
ra1l [238]

Answer:

Flow rate is 1.82\times 10^{-8} m^{3}/s

Explanation:

Given information

Density of oil, \rho_{oil}= 850 Kg/m^{3}

kinematic viscosity, v= 0.00062 m^{2} /s

Diameter of pipe, D= 5 mm= 0.005 m

Length of pipe, L=40 m

Height of liquid, h= 3 m

Volume flow rate for horizontal pipe will be given by

\bar v=\frac {\triangle P\pi D^{4}}{128\mu L} where \mu is dynamic viscosity and \triangle P is pressure drop

At the bottom of the tank, pressure is given by

P_{bottom}=\rho_{oil} gh=850 Kg/m^{3}\times 9.81 m/s^{2}\times 3 m= 25015.5 N/m^{2}

Since at the top pressure is zero, therefore change in pressure a difference between the pressure at the bottom and the top. It implies that change in pressure is still 25015.5 N/m^{2}

Dynamic viscosity, \mu=\rho_{oil}v= 850 Kg/m^{3}\times 0.00062 m^{2}/s=0.527 Kg/m.s

Now the volume flow rate will be

\bar v=\frac {25015.5 N/m^{2}\times \pi \times 0.005^{4}}{128\times 0.527 Kg/m.s \times 40}=1.82037\times 10^{-8} m^{3}/s\approx 1.82\times 10^{-8} m^{3}/s

Proof of flow being laminar

The velocity of flow is given by

V_{flow}=\frac {\bar v}{A}=\frac {1.82\times 10^{-8} m^{3}/s}{0.25\times \pi\times 0.005^{2}}=0.000927104  m/s

Reynolds number, Re=\frac {\rho_{oil} v_{flow} D}{\mu}=\frac {850 Kg/m^{3}\times 0.000927104 m/s\times 0.005}{0.527 kg/m.s}=0.007476648

Since the Reynolds number is less than 2300, the flow is laminar and the assumption is correct.

5 0
3 years ago
A 24-volt DC battery is connected
butalik [34]
The answer will be : 0 volts.
4 0
2 years ago
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