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ser-zykov [4K]
3 years ago
7

Any help is appreciated.

Engineering
1 answer:
xenn [34]3 years ago
4 0
The answer to this question is true
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Air flows at 45m/s through a right angle pipe bend with a constant diameter of 2cm. What is the overall force required to keep t
HACTEHA [7]

Answer:

b)1.08 N

Explanation:

Given that

velocity of air V= 45 m/s

Diameter of pipe = 2 cm

Force exerted by fluid  F

F=\rho AV^2

So force exerted in x-direction

F_x=\rho AV^2

F_x=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2

F=0.763 N

So force exerted in y-direction

F_y=\rho AV^2

F_y=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2

F=0.763 N

So the resultant force R

R=\sqrt{F_x^2+F_y^2}

R=\sqrt{0.763^2+0.763^2}

R=1.079

So the force required to hold the pipe is 1.08 N.

3 0
3 years ago
Technician A says a sway bar is used to control body lean when cornering.
vodka [1.7K]

Answer:

technician B

Explanation:

because adding leaves increasing capacity

6 0
3 years ago
The first thing you are going to create is a logical one bit full adder in continuous assignment verilog. You can only use logic
IgorC [24]

Answer:

See Explaination

Explanation:

// use the `timescale directive which u have used in ur testbench here

module FA1(a,b,cin,s,cout);

input a,b,cin;

output s,cout;

wire s1,c1,c2;

assign s1= #4 a ^ b;

assign s= #4 s1 ^ cin;

assign c1= #2 a & b;

assign c2= #2 s1 & cin;

assign cout= #3 c1 | c2;

endmodule

5 0
3 years ago
Reaming is used for which three of the following functions: (a) accurately locate a hole position, (b) create a stepped hole, (c
Svetlanka [38]

Answer:

b

Explanation:

4 0
3 years ago
You have been assigned the task of reviewing the relief scenarios for a specific chemical reactor in your plant. You are current
postnew [5]

Answer:

D=0.016m

Explanation:

From the question we are told that:

Discharge Rate F_r=0.5kgls

Pressure P=15Kpa

Temperature T=25=>298K

Ambient pressure is 1 atm.

Generally the equation for Density is mathematically given by

\rho=\frac{PM}{RT}

\rho=\frac{15*10^5*28.0134*10^{-3}}{8.314*298}

\rho=16.958kg/m^2

Generally the equation for Flow rate is mathematically given by

F_r=\mu A\sqrt{Q \rho P(\frac{2}{Q+1})^{\frac{Q+1}{Q-1}}}

Where

Q=Heat coefficient\ ratio\ of\ Nitrogen

Q=1.4

\mu= Discharge\ coefficient

\mu=0.68

Therefore

0.5=0.68 A\sqrt{1.4 16.958 15*10^{5}(\frac{2}{1.4+1})^{\frac{1.4+1}{1.4-1}}}

A=2.129*10^{-4}

Where

A=\frac{\pi}{4}D^2

\frac{\pi}{4}D^2=2.129*10^{-4}

D=0.016m

8 0
3 years ago
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