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Anuta_ua [19.1K]
3 years ago
12

I have a 308 bullet stuck in my ak,s mag help?

Engineering
2 answers:
polet [3.4K]3 years ago
4 0
Why not just use slight oil(olive oil or Coconut oil)... so you’re not getting anything on your tongue from the bullet?
Assoli18 [71]3 years ago
3 0
Lick the bullet and push it down
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The undisturbed soil at given borrow pit is found to have the following property:
tankabanditka [31]

Answer:

Check the explanation

Explanation:

Determine the weight o ids in the each truck °slim the relation,  

W,=\frac{W}{1+w}

Here, W is net weight of soil and water on the truck and w is water content  

substitute 72 7 kN for W and 15% for w.

you will need to also determine the number of truck loads required using the relation:

Number of truck loads required = \frac{W_{sc} }{W_{s}}

Kindly check the attached image below for the full explanation to the question above.

7 0
3 years ago
Select three ways an engineer can create a view of a design.
Angelina_Jolie [31]

Sketching projection

  • Sketching will give the brief planning of the engineering work to be done so it's important

Isometric projection

  • Yes doing a work in 3D factor out the mistakes which can be optimised

Third angle projection:-

  • Viewing angles matter a lot
  • So it will help the engineer to make it perfect from all angles
5 0
2 years ago
Design the software architecture for a student online registration system using the MVC architecture. Draw the architecture by s
Travka [436]

MVC architecture is defined as the architectural design that is used by software engineers for programming languages.

<h3>What are the various models of MVC architecture?</h3>

The various types of MVC architecture include the following:

  • The controller: This model is used to control logic and acts as the coordinator between the View and the Model.

  • The view: It displays the information from the model to the user.

  • The model: It is used to implement the domain logic.

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8 0
2 years ago
A cylindrical bar of metal having a diameter of 17.8 mm and a length of 196 mm is deformed elastically in tension with a force o
ExtremeBDS [4]

Answer:

A) ΔL = 0.544 mm

B) Δd = -0.0168 mm

Explanation:

We are given;

Elastic modulus; E = 67.1 GPa = 67.1 × 10^(9) Pa

Force; F = 46400 N

Diameter; d = 17.8 mm = 17.8 × 10^(-3) m

Radius; r = d/2 = 17.8/2 = 8.9 mm = 8.9 × 10^(-3) m

Length; L = 196 mm = 0.196 m

Poisson ratio; ν = 0.34

A) We know that formula for elastic modulus is;

E = σ/ε

Where;

σ = F/A

ε = ΔL/L

Thus;

E = (FL/ΔL•A)

ΔL is change in length. Making it the subject of the formula, we have;

ΔL = FL/AE

Now, A = πr²

A = π × (8.9 × 10^(-3))²

ΔL = [(0.196 × 46400)/(π × (8.9 × 10^(-3))² × 67.1 × 10^(9)]

ΔL = 0.544 × 10^(-3) m

ΔL = 0.544 mm

B) formula for Poisson ratio is given as;

ν = -ε_x/ε_z

Where;

ε_x is transverse strain = Δd/d

ε_z is longitudinal strain = ΔL/L

Thus;

ν = -Δd•L/d•ΔL

Making Δd the subject, we have;

Δd = -νdΔL/L

Δd = -(0.34 × 17.8 × 10^(-3) × 0.544 × 10^(-3))/0.196

Δd = -0.0168 × 10^(-3) m = -0.0168 mm

7 0
3 years ago
1. 6.1 PSPICEMULTISIM The current in a 50μH inductor is known to be iL=18te−10tAfor t≥0. 1. Find the voltage across the inductor
Andre45 [30]

Answer:

a. Voltage across the inductor for t > 0 is 0.9e^-10t(1-10t)

b. Power = -59.3μW

c. Inductor is delivering power.

d. Energy = 5934.3nJ

e. Time = 100ms; Energy = 1095941.025nJ

Explanation:

Given

Current; iL=18te^(−10t)A or t≥0.

L.= inductor = 50μH

a. The voltage, V across the inductor for t>0 is calculated as follows;

V = L(di/dt)

Where L = 50μH

di/dt = 18(e^-10t + (-10)te^-10t)

di/dt = 18e^-10t(1 - 10t)

Substitute 50μH for L and 18e^-10t(1 - 10t) for di/dt in V = L(di/dt)

V = 50μH * 18e^-10t(1 - 10t)

V = 50 * 10^-6(18e^-10t(1 - 10t))

V = 0.9e^-10t(1-10t)

Hence, the voltage across the inductor for t > 0 is 0.9e^-10t(1-10t)

b. Find the power (in microwatts) at the terminals of the inductor when t=200 ms.

Given that t = 200ms = 200 * 10^-3s = 0.2s

Power, p is calculated using the following formula;

p = Li(di/dt)

p = 50 * 10^-6(18te^-10t)18e^-10t(1-10t)

p = 50 * 10^-6 * (18 * 0.2 * e^-(10*0.2)) * (18 * e^(-10 * 0.2) * (1-10*0.2)

p = -5.93E5W

p = -59.3μW

c. Is the inductor absorbing or delivering power at 200 ms?

Because of the negative sign, the inductor is delivering power.

d. Find the energy (in microjoules) stored in the inductor at 200 ms.

Energy is calculated as ½Li²

= ½ * 50 * 10^-6 * (18te^-10t)²

= ½ * 50 * 10^-6 * (18 * 0.2 * e ^ (-10 * 0.2))²

= 0.0000059342669999498J

= 5934.3nJ

e. Find the maximum energy (in microjoules) stored in the inductor and the time (in milliseconds) when it occurs.

Calculating the derivation in (a)

di/dt = 0

18e^-10t(1-10t) = 0

1 - 10t = 0

-10t = -1

t = 1/10

t = 100ms

To calculate the energy, first we need to calculate the current

I(t=100) = 18 * 0.1 * e^(-10(0.1)

I = 0.662182994108596

I = 6621.82mA

The energy is calculated as follows;

w = ½ * 50 * 10^-6 * (6.621)²

w = 0.001095941025

w = 1095941.025nJ

8 0
3 years ago
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