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2 years ago

You are traveling upstream on a river at dusk. You see a buoy with the number 5 and a flashing green light . What should you do?

Engineering

1 answer:

nalin [4]2 years ago

8 0

Answer:

please give brainliest my brother just got the corona virus

Explanation:

this is my brothers account he wants to get 5 brainliest

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The stagnation chamber of a wind tunnel is connected to a high-pressure airbottle farm which is outside the laboratory building.

Natasha2012 [34]

This question is not complete, the complete question is;

The stagnation chamber of a wind tunnel is connected to a high-pressure air bottle farm which is outside the laboratory building. The two are connected by a long pipe of 4-in inside diameter. If the static pressure ratio between the bottle farm and the stagnation chamber is 10, and the bottle-farm static pressure is 100 atm, how long can the pipe be without choking? Assume adiabatic, subsonic, one-dimensional flow with a friction coefficient of 0.005

Answer:

the length of the pipe is 11583 in or 965.25 ft

Explanation:

Given the data in the question;

Static pressure ratio; p1/p2 = 10

friction coefficient f = 0.005

diameter of pipe, D =4 inch

first we obtain the value from FANN0 FLOW TABLE for pressure ratio of ( p1/p2 = 10 )so

4fL $_{max}$ / D = 57.915

we substitute

(4×0.005×L $_{max}$ ) / 4 = 57.915

0.005L $_{max}$ = 57.915

L $_{max}$ = 57.915 / 0.005

L $_{max}$ = 11583 in

Therefore, the length of the pipe is 11583 in or 965.25 ft

6 0

2 years ago

the velocity of a particle is given by v=16t^2i+4t^3j+(5t+2k) m/s, where t is in seconds. if the particle is at the origin when

pshichka [43]

Answer:

80.16 m/s^2

at t=2 s

x=42.3 m

y=16 m

z=14 m

Explanation:

solution

The x,y,z components of the velocity are donated by the i,j,k vectors.

$v_{x}=16t^{2} \\v_{y}=4t^{3}\\v_{z}=5t+2$

acceleration is a derivative of velocity with respect to time.

$a_{x}=\frac{d}{dt} v_{x}=\frac{d}{dt}[16t^{2}]=32t\\a_{y}=\frac{d}{dt} v_{y}=\frac{d}{dt}[4t^{3}]=12t^{2} \\a_{z}=\frac{d}{dt} v_{z}=\frac{d}{dt}[5t+2]=5$

evaluate acceleration at 2 seconds

$a_{x} =32*2=64m/s^{2}\\ a_{y} =12*2^{2} =48m/s^{2}\\a_{z} =5m/s^{2}$

the magnitude of the acceleration is the square root of the sum of the square of each component of the acceleration.

$=\sqrt{a_{x}^2 +a_{y}^2+a_{z} ^2 } \\=\sqrt{64^2 +48^2+5 ^2 }\\=80.16m/s^2$

position is the integral of velocity with respect to time position at a time can be found by taking by taking the definite intergral of each component.

$x=\int\limits {v_{x} } \, dx=\int\limits^2_0 {{16t^2} \, dt=42.7m\\\\y=\int\limits {v_{y} } \, dx=\int\limits^2_0 {{4t^3} \, dt=16m\\\\\\\\\\z=\int\limits {v_{z} } \, dx=\int\limits^2_0 {{5t+2} \, dt=14m\\\\$