All categories

3 years ago

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.

You might be interested in

An air-standard dual cycle has a compression ratio of 9.1 and displacement of Vd = 2.2 L. At the beginning of compression, p1 =

jok3333 [9.3K]

Answer:

a) T₂ is 701.479 K

T₃ is 1226.05 K

T₄ is 2350.34 K

T₅ is 1260.56 K

b) The net work of the cycle in kJ is 2.28 kJ

c) The power developed is 114.2 kW

d) The thermal efficiency, $\eta _{dual}$ is 53.78%

e) The mean effective pressure is 1038.25 kPa

Explanation:

a) Here we have;

$\frac{T_{2}}{T_{1}}=\left (\frac{v_{1}}{v_{2}} \right )^{\gamma -1} = \left (r \right )^{\gamma -1} = \left (\frac{p_{2}}{p_{1}} \right )^{\frac{\gamma -1}{\gamma }}$

Where:

p₁ = Initial pressure = 95 kPa

p₂ = Final pressure =

T₁ = Initial temperature = 290 K

T₂ = Final temperature

v₁ = Initial volume

v₂ = Final volume

$v_d$ = Displacement volume =

γ = Ratio of specific heats at constant pressure and constant volume cp/cv = 1.4 for air

r = Compression ratio = 9.1

Total heat added = 4.25 kJ

1/4 × Total heat added = $c_v \times (T_3 - T_2)$

3/4 × Total heat added = $c_p \times (T_4 - T_3)$

$c_v$ = Specific heat at constant volume = 0.718×2.821× 10⁻³

$c_p$ = Specific heat at constant pressure = 1.005×2.821× 10⁻³

v₁ - v₂ = 2.2 L

$\left \frac{v_{1}}{v_{2}} \right =r \right = 9.1$

v₁ = v₂·9.1

∴ 9.1·v₂ - v₂ = 2.2 L = 2.2 × 10⁻³ m³

8.1·v₂ = 2.2 × 10⁻³ m³

v₂ = 2.2 × 10⁻³ m³ ÷ 8.1 = 2.72 × 10⁻⁴ m³

v₁ = v₂×9.1 = 2.72 × 10⁻⁴ m³ × 9.1 = 2.47 × 10⁻³ m³

Plugging in the values, we have;

${T_{2}}= T_{1} \times \left (r \right )^{\gamma -1} = 290 \times 9.1^{1.4 - 1} = 701.479 \, K$

From;

$\left (\frac{p_{2}}{p_{1}} \right )^{\frac{\gamma -1}{\gamma }}= \left (r \right )^{\gamma -1}$ we have;

$p_{2} = p_{1}} \times \left (r \right )^{\gamma } = 95 \times \left (9.1 \right )^{1.4} = 2091.13 \ kPa$

1/4×4.25 = $0.718 \times 2.821 \times 10^{-3}\times (T_3 - 701.479)$

∴ T₃ = 1226.05 K

Also;

3/4 × Total heat added = $c_p \times (T_4 - T_3)$ gives;

3/4 × 4.25 = $1.005 \times 2.821 \times 10^{-3} \times (T_4 - 1226.05)$ gives;

T₄ = 2350.34 K

$\frac{T_{4}}{T_{5}}=\left (\frac{v_{5}}{v_{4}} \right )^{\gamma -1} = \left (\frac{r}{\rho } \right )^{\gamma -1}$

$\rho = \frac{T_4}{T_3} = \frac{2350.34}{1226.04} = 1.92$

$T_{5} = \frac{T_{4}}{\left (\frac{r}{\rho } \right )^{\gamma -1}}= \frac{2350.34 }{\left (\frac{9.1}{1.92 } \right )^{1.4-1}} =1260.56 \ K$

b) Heat rejected = $c_v \times (T_5 - T_1)$

$Therefore \ heat \ rejected = 0.718 \times 2.821 \times 10^{-3}\times (1260.56 - 290) = 1.966 kJ$

The net work done = Heat added - Heat rejected

∴ The net work done = 4.25 - 1.966 = 2.28 kJ

The net work of the cycle in kJ = 2.28 kJ

c) Power = Work done per each cycle × Number of cycles completed each second

Where we have 3000 cycles per minute, we have 3000/60 = 50 cycles per second

Hence, the power developed = 2.28 kJ/cycle × 50 cycle/second = 114.2 kW

$Thermal \ efficiency, \, \eta _{dual} = \frac{Work \ done}{Heat \ supplied} = \frac{2.28}{4.25} \times 100 = 53.74 \%$

The thermal efficiency, $\eta _{dual}$ = 53.78%

e) The mean effective pressure, $p_m$ , is found as follows;

$p_m = \frac{W}{v_1 - v_2} =\frac{2.28}{2.2 \times 10^{-3}} = 1038.25 \ kPa$

The mean effective pressure = 1038.25 kPa.

3 0

4 years ago

This elementary problem begins to explore propagation delayand transmission delay, two central concepts in data networking. Cons

telo118 [61]

Explanation:

(a)

Here, distance between hosts A and B is m meters and, propagation speed along the link is s meter/sec

Hence, propagation delay, $d_{prop} = m/sec (s)$

(b)

Here, size of the packet is L bits

And the transmission rate of the link is R bps

Hence, the transmission time of the packet, $d_{trans} = L/R$

(c)

As we know, end-to-end delay or total no delay,

$\mathrm{d}_{\text {nodal }}=\mathrm{d}_{\text {proc }}+\mathrm{d}_{\text {quar }}+d_{\max }+d_{\text {prop }}$

$Here, $\mathrm{d}_{\text {rroc }}$ and $\mathrm{d}_{\text {quat }}$ \\Hence, $\mathrm{d}_{\text {rodal }}=\mathrm{d}_{\text {trass }}+\mathrm{d}_{\text {prop }}$ \\We know, $\mathrm{d}_{\text {trax }}=\mathrm{L} / \mathrm{R}$ sec and $\mathrm{d}_{\text {vapp }}=\mathrm{m} / \mathrm{s}$ sec$ $\text { Hence, } {d_{\text {nodal }}}=\mathrm{L} / \mathrm{R}+\mathrm{m} / \mathrm{s} \text { seconds }$

(d)

The expression, time $time $t=d_{\text {trans }}$$ means the\at time since transmission started is equal to transmission delay.

As we know, transmission delay is the time taken by host to push out the packet.

Hence, at $time $t=d_{\text {trans }}$$ the last bit of the packet has been pushed out or transmitted.

(e)

If $\ d_{prop} >d_{trans}$

Then, at $time $t=d_{\text {trans }}$$ the bit has been transmitted from host A, but to condition (1), the first bit has not reached B.

(f)

If $\ d_{prop}$

Then, at $time $t=d_{\text {trans }}$$ , the first bit has reached destination on B

Here, $s=2.5 \times 10^{8} \mathrm{sec}$

$\begin{aligned}&\mathrm{L}=100 \mathrm{Bits} \text { and }\\&\mathrm{R}=28 \mathrm{kbps} \text { or } 28 \times 1000 \mathrm{bps}\end{aligned}$

It's given that $\ d_{prop} =d_{trans}$

Hence,

$\begin{aligned}\ & \frac{L}{R}=\frac{m}{s} \\m &=s \frac{L}{R} \\&=\frac{2.5 \times 10^{8} \times 100}{28 \times 1000} \\&=892.9 \mathrm{km}\end{aligned}$

5 0

3 years ago

Rod of steel, 200 mm length reduces its diameter (50 mm) by turning by 2 mm with feed speed 25 mm/min. You are required to calcu

diamong [38]

Answer:

125 cm³/min

Explanation:

The material rate of removal is usually given by the formula

Material Rate of Removal = Radial Depth of Cut * Axial Depth of Cut * Feed Rate, where

Radial Depth of Cut = 25 mm

Axial depth of cut = 200 mm

Feed rate = 25 mm/min

On multiplying all together, we will then have

MRR = 25 mm * 200 mm * 25 mm/min

MRR = 125000 mm³/min

Or we convert it to cm³/min and have

MRR = 125000 mm³/min ÷ 1000

MRR = 125 cm³/min

4 0

3 years ago

Why is the contractor normally required to submit a bid bond when making a proposal to an owner on a competitively bid contract?

just olya [345]

Answer:

It serves as a guarantee that the contractor who wins the bid will honor the terms of the bid after the contract is signed.

Explanation:

A bid bond is a type of construction bond that protects the obligee in a construction bidding process.

A bid bond typically involves three parties:

The obligee; the owner or developer of the construction project under bid. The principal; the bidder or proposed contractor.

The surety; the agency that issues the bid bond to the principal example insurance company or bank.

A bid bond generally serves as a guarantee that the contractor who wins the bid will honor the terms of the bid after the contract is signed.

3 0

3 years ago

In the early 1900s what was developed and used that opened the door for naval use of aircraft

Alona [7]

Answer:

ww1

Explanation:

5 0

4 years ago

Read 2 more answers

How an AK 47 gun was works​

How an AK 47 gun was works