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2 years ago

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

Engineering

2 answers:

polet [3.4K]2 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]2 years ago

3 0

Lick the bullet and push it down

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The ice on the rear window of an automobile is defrosted by attaching a thin, transparent, film type heating element to its inne

pshichka [43]

Answer:

A)Q = 1208.33 W/m²

B)K = 0.138 W/m.K

Explanation:

We are given;

inside air temperature;T_∞,i =25 °C = 25 + 273 = 298K

outside air temperature;T_∞,o = -10°C = - 10 + 273 = 263K

Inner surface temperature;T_s,i = 15 °C = 15 + 273 = 288K

Thickness, L = 4mm = 0.004m

convection heat transfer coefficient ; hi = 25 W/(m².K)

A) From an energy balance at the inner surface and the thermal circuit, the electric power required per unit window area is given as;

Q = [(T_s,i - T_∞,o)/((L/k) + (1/hi))] - [(T_∞,o - T_s,i)/(1/hi)]

Plugging in the relevant values with k for glass as 1.4 W/m.k, we have;

Q = [(288 - 263)/((0.004/1.4) + (1/25))] - [(263 - 288)/(1/25)]

Q = 583.33 + 625

Q = 1208.33 W/m²

B) The formula for thermal conductivity is;

K = (QL)/(AΔT)

Where;

K is the thermal conductivity in W/m.K

Q is the amount of heat transferred through the material

L is the distance between the two isothermal planes

A is the area of the surface in square meters

ΔT is the difference in temperature in Kelvin

ΔT = 298K - 263K = 35K

Now, since we have value of heat per unit area to be Q = 1208.33 W/m², let's rearrange the equation to reflect that; Thus ;

k = (Q/A) x (L/ΔT)

K = 1208.33 x (0.004/35)

K = 0.138 W/m.K

5 0

2 years ago

A plate and frame heat exchanger has 15 plates made of stainless steel that are 1 m tall. The plates are 1 mm thick and 0.6 m wi

hodyreva [135]

Answer:

14.506°C

Explanation:

Given data :

flow rate of water been cooled = 0.011 m^3/s

inlet temp = 30°C + 273 = 303 k

cooling medium temperature = 6°C + 273 = 279 k

flow rate of cooling medium = 0.02 m^3/s

Determine the outlet temperature

we can determine the outlet temperature by applying the relation below

Heat gained by cooling medium = Heat lost by water

= ( Mcp ( To - 6 ) = Mcp ( 30 - To )

since the properties of water and the cooling medium ( water ) is the same

= 0.02 ( To - 6 ) = 0.011 ( 30 - To )

= 1.82 ( To - 6 ) = 30 - To

hence To ( outlet temperature ) = 14.506°C

6 0

2 years ago

A race car is travelling around a circular track. The velocity of the car is 20 m/s, the radius of the track is 300 m, and the m

Zielflug [23.3K]

Answer:

μ = 0.136

Explanation:

given,

velocity of the car = 20 m/s

radius of the track = 300 m

mass of the car = 2000 kg

centrifugal force

$F_c = \dfrac{mv^2}{r}$

$F_c = \dfrac{2000\times 20^2}{300}$

F c = 2666. 67 N

F f= μ N

F f = μ m g

2666.67 = μ × 2000 × 9.8

μ = 0.136

so, the minimum coefficient of friction between road surface and car tyre is equal to μ = 0.136

5 0

3 years ago

Air at T1 = 32°C, p1 = 1 bar, 50% relative humidity enters an insulated chamber operating at steady state with a mass flow rate

RUDIKE [14]

Answer:

4.5kg/min

Explanation:

Given parameters

$T_1 = 32^0 C, m_1 = 3 kg/min, T_2 = 7^0 C ,T_3 = 17^0$

if we take

The mass flow rate of the second stream = $m_2(kg/min)$

The mass flow rate of mixed exit stream = $m_3 (kg/min)$

Now from mass conservation

$m_3 = m_2 + m_1$

$m_3 = m_2 + 3 (kg/min)$

The temperature of the mixed exit stream given as

$T_3m_3 = T_2m_2 +T_1m_1\\\\17 ( 3 + m_2) = 7 \times m_2 + 32 \times 3\\\\51 + 17 m_2 = 7 m_2 + 96\\\\10 m_2 = 96 - 51\\\\m_2 = 4.5 kg/min\\\\\\\\$

Therefore the mass flow rate of second stream will be 4.5 kg/min.

7 0

2 years ago

A cone-shaped part is to be fabricated using stereolithography. The radius of the cone at its base = 35 mm, and its height = 50

iren [92.7K]

Explanation:

volume = πR²h/3

= πx35²x50/3

= 192325/3

= 64166.725

number of layers n = 50mm/0.05

n = 1000layers

average volume = 64166.725/1000

= 64.167mm³

average area = 64.167/0.05

= 1283.34mm²

average time = 1283.34/900mm x 0.22

= 6.48

6.48 + 15 seconds

= 21.48 seconds

time required = 1000x21.48

= 21480 seconds

convert to minutes

21480/60

= 358 minutes

21480/3600

= 5.967 hours

8 0

2 years ago