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Svetllana [295]
3 years ago
15

I JUST PLAYED A GAME IN 2ND PERSON AND IM CONFEUSED

Engineering
2 answers:
sattari [20]3 years ago
6 0

Why would you be confused

Also which game

Crazy boy [7]3 years ago
3 0

Answer:

cool

Explanation:

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Knowing that the central portion of the link BD has a uniform cross-sectional area of 800 2, determine the magnitude of the load
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A laboratory in the Y building keep a vacuum pressure of 0.1 kPa abs. What is the net force acting on the door considering the a
seropon [69]

Answer:

net force acting on the floor is 100 kN

Explanation:

Given data:

P_{vaccum} = 0.1 kPa

P_{atm} = 101.325 kPa

dimension of floor = 2 m \times 0.5 m

we know that

Net force can be calculated as follow

f_{net} = P_{vaccum} \times area

f_{net} = 0.1\times 10^3 \times 2\times 0.5

f_{net} = 0.1\times 10^3 \times 1

f_{net} = 100 kN

Therefore net force acting on the floor is 100 kN

7 0
4 years ago
A 2-lane highway is to be constructed across a 6-ft diameter metal culvert which is oriented perpendicular to the highway center
kvv77 [185]

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Dude just like forget the highway and drive on the interstate bruh

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5 0
4 years ago
A thick casting with a thermal diffusivity of 5 x 10-6 m2/s is initially at a uniform temperature of 150oC. One surface of the c
fredd [130]

Answer:

T_o = 141.81 ^0C

Explanation:

Given that;

Thermal diffusivity \alpha = 5 \times 10 ^{-6} m^2/s

Thermal conductivity k = 20 \ W/m.K

Heat transfer coefficient h = ( we are to assume the imposed surface temperature ) = 20 W/m².K

Initial temperature = 150 ° C = (150+273) K = 423 K

Then coolant temperature with which the casting is exposed to = 20° C = (20+273)K = 293 K

Time = 40 seconds

Length = 20mm = 0.02 m

The objective is to determine the  temperature at the surface  at a depth of 20 mm after 40 seconds.

Bi = \dfrac{hL}{k}

Bi = \dfrac{20*0.02}{20}

Bi == 0.02

\tau = \dfrac{\alpha t}{L^2}

\tau=  \dfrac{5*10^{-6 }* 40}{0.020^2}

\tau = 0.5

For a wall at 0.2 Bi

A_1 = 1.0311

\lambda _1 = 0.4328

Therefore;

\dfrac{T_o - T_{\infty}}{T_i - T_{\infty}}= A_1 e ^{-( \lambda_1^2 \ \tau)

\dfrac{T_o - 293 }{423 - 293}= 1.0311 \times e ^{-( 0.438^2 \times 0.5 )

\dfrac{T_o - 293 }{423 - 293}= 1.0311 \times e ^{-( 0.0959 )

\dfrac{T_o - 293 }{130}= 1.0311 \times 0.9085

\dfrac{T_o - 293 }{130}= 0.937

T_o - 293= 0.937 \times 130

T_o - 293= 121.81

T_o = 121.81+ 293

T_o = 414.81 \ K

T_o = (414.81 - 273)^0C

T_o = 141.81 ^0C

4 0
4 years ago
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