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andrew11 [14]
3 years ago
14

How an AK 47 gun was works​

Engineering
1 answer:
monitta3 years ago
4 0

Explanation:

Basically the operator pulls the trigger, which releases the firing hammer, and strikes the firing pin making the gunpowder ignites.

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(20 points) A 1 mm diameter tube is connected to the bottom of a container filled with water to a height of 2 cm from the bottom
zzz [600]

Solution :

Given :

h = 2 cm

Diameter of the tube , d = 1 mm

Diameter of the hose, D = 6 mm

Between 1 and 2, by applying Bernoulli's principle, we get

As point 1 is just below the free surface of liquid, so

$P_1=P_{atm} \text{ and} \ V_1=0$

$\frac{P_{atm}}{\rho g}+\frac{v_1^2}{2g} +h = \frac{P_2}{\rho g}$

$\frac{101.325}{1000 \times 9.81}+0.02 =\frac{P_2}{\rho g}$

$P_2 = 111.35 \ kPa$

Therefore, 111.325 kPa is the gas supply pressure required to keep the water from leaking back into the tube.

Velocity at point 2,

$V_2=\sqrt{\left(\frac{111.135}{\rho g}+0.02}\right)\times 2g$

   = 1.617 m/s

Flow of water,  $Q_2 = A_{tube} \times V_2$

                               $=\frac{\pi}{4} \times (10^{-3})^2 \times 1.617 $

                               $1.2695 \times 10^{-6} \ m^3/s$

Minimum air flow rate,

$Q_2 = Q_3 = A_{hose} \times V_3$

$V_3 = \frac{Q_2}{\frac{\pi}{4}D^2}$

$V_3 = \frac{1.2695 \times10^{-6}}{\pi\times 0.25 \times 36 \times 10^{-6}}$

    = 0.0449 m/s

b). Reynolds number in hose,

$Re = \frac{\rho V_3 D}{\mu} = \frac{V_3 D}{\nu}$

υ for water at 25 degree Celsius is $8.9 \times 10^{-7} \ m^2/s$

υ for air at 25 degree Celsius is $1.562 \times 10^{-5} \ m^2/s$

$Re_{hose}=\frac{0.0449 \times 6 \times 10^{-3}}{1.562 \times 10^{-5}}$

           = 17.25

Therefore the flow is laminar.

Reynolds number in the pipe

$Re = \frac{V_2 d}{\nu} = \frac{1.617 \times 10^{-3}}{8.9 \times 10^{-7}}$

                = 1816.85, which is less than 2000.

So the flow is laminar inside the tube.

3 0
2 years ago
(a) Determine the dose (in mg/kg-day) for a bioaccumulative chemical with BCF = 103 that is found in water at a concentration of
solmaris [256]

Answer:

0.064 mg/kg/day

6.25% from water, 93.75% from fish

Explanation:

Density of water is 1 kg/L, so the concentration of the chemical in the water is 0.1 mg/kg.

The BCF = 10³, so the concentration of the chemical in the fish is:

10³ = x / (0.1 mg/kg)

x = 100 mg/kg

For 2 L of water and 30 g of fish:

2 kg × 0.1 mg/kg = 0.2 mg

0.030 kg × 100 mg/kg = 3 mg

The total daily intake is 3.2 mg.  Divided by the woman's mass of 50 kg, the dosage is:

(3.2 mg/day) / (50 kg) = 0.064 mg/kg/day

b) The percent from the water is:

0.2 mg / 3.2 mg = 6.25%

And the percent from the fish is:

3 mg / 3.2 mg = 93.75%

3 0
3 years ago
Determine the total condensation rate of water vapor onto the front surface of a vertical plate that is 10 mm high and 1 m in th
castortr0y [4]

Answer:

Q =  63,827.5 W

Explanation:

Given:-

- The dimensions of plate A = ( 10 mm x 1 m )

- The fluid comes at T_sat , 1 atm.

- The surface temperature, T_s = 75°C  

Find:-

Determine the total condensation rate of water vapor onto the front surface of a vertical plate

Solution:-

- Assuming drop-wise condensation the heat transfer coefficient for water is given by Griffith's empirical relation for T_sat = 100°C.

                            h = 255,310 W /m^2.K

- The rate of condensation (Q) is given by Newton's cooling law:

                           Q = h*As*( T_sat - Ts )

                           Q = (255,310)*( 0.01*1)*( 100 - 75 )

                           Q =  63,827.5 W

8 0
3 years ago
Read 2 more answers
I need your help to answer in the picture plss help me!​
Aloiza [94]

this looks like its the different phases of a single cylinder 4 stroke engine what are you doing in the picture or assignment though matching the numbers to the descriptions on the side?

3 0
3 years ago
Balanced forces will result in which of the following
tia_tia [17]
D is the answer. Hope this helped
8 0
3 years ago
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