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2 years ago

PLEASE HELP AND ANSWER MY OTHER QUESTIONS!,

Engineering

1 answer:

NARA [144]2 years ago

8 0

Answer: False

Explanation:

The statement that "a person hired by a pharmaceutical company to streamline the company’s drug production process would most likely be an electrical engineer" is false.

An electrical engineer is an engineer that designs, and tests electrical equipment like smartphones, electric motors, etc. The person designs equipments which makes use of electricity or electronics.

He or she doesn't streamline the company’s drug production process, this can be the role played by the production manager.

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Refrigerant-134a enters a 28-cm-diameter pipe steadily at 200 kPa and 20°C with a velocity of 5 m/s. The refrigerant gains heat

Alexandra [31]

Answer:

V = 0.30787 m³/s

m = 2.6963 kg/s

v2 = 0.3705 m³/s

v2 = 6.017 m/s

Explanation:

given data

diameter = 28 cm

steadily =200 kPa

temperature = 20°C

velocity = 5 m/s

solution

we know mass flow rate is

m = ρ A v

floe rate V = Av

m = ρ V

flow rate = V = $\frac{m}{\rho}$

V = Av = $\frac{\pi}{4} * d^2 * v1$

V = $\frac{\pi}{4} * 0.28^2 * 5$

V = 0.30787 m³/s

and

mass flow rate of the refrigerant is

m = ρ A v

m = ρ V

m = $\frac{V}{v}$ = $\frac{0.30787}{0.11418}$

m = 2.6963 kg/s

and

velocity and volume flow rate at exit

velocity = mass × v

v2 = 2.6963 × 0.13741 = 0.3705 m³/s

and

v2 = A2×v2

v2 = $\frac{v2}{A2}$

v2 = $\frac{0.3705}{\frac{\pi}{4} * 0.28^2}$

v2 = 6.017 m/s

7 0

3 years ago

Question 21(Multiple Choice

Katen [24]

Answer:

four seconds

Explanation:

lookin at a vehicle respectively at a second can cause accident

8 0

2 years ago

Read 2 more answers

An ideal gas initially at 300 K and 1 bar undergoes a three-step mechanically reversible cycle in a closed system. In step 12, p

Veseljchak [2.6K]

Answer:

Ts =Ta E)- 300(

569.5 K

Q12-W12 = -4014.26

Mol

AU2s = Q23= 5601.55

Mol

AUs¡ = Ws¡ = -5601.55

Explanation:

A clear details for the question is also attached.

(b) The P,V and T for state 1,2 and 3

P =1 bar Ti = 300 K and Vi from ideal gas Vi=

24.9x10 m

P-5 bar

Due to step 12 is isothermal: T1 = T2= 300 K and

VVi24.9 x 10x-4.9 x 10-3 *

The values at 3 calclated by Uing step 3l Adiabatic process

B-P ()

Since step 23 is Isochoric: Va =Vs= 4.99 m* and 7=

Ps-1x(4.99 x 103

P-1x(29x 10)

9.49 barr

And Ts =Ta E)- 300(

569.5 K

Work done for Isotermal process define as

8.314 x 300 In =4014.26

Wi2= RTi ln

mol

And fromn first law of thermodynamic

AU12= W12 +Q12

Q12-W12 = -4014.26

Mol

F'or step 23 Isochoric: AV = 0 Since volume change is zero W23= 0 and

Alls = Cp(L3-12)=5 x 8.311 (569.5 - 300) = 7812.18-

AU23= C (13-72) =5 x 8.314 (569.3 - 300) = 5601.53

Inol

Now from first law of thermodynamic the Q23

AU2s = Q23= 5601.55

Mol

For step 3-1 Adiabatic: Since in this process no heat transfer occur Q31= 0

and

C,(T -Ts)=x 8.314 (300- 569.5)= -7842.18

mol

AU=C, (T¡-T)= x 8.314 (300

-5601.55

569.5)

mol

Now from first law of thermodynamie the Ws1

mol

AUs¡ = Ws¡ = -5601.55

3 0

2 years ago

Find the number of Btu conducted through a wall in 8 hours. The wall is 8 feet high by 24 feet long and has a total R-value of 1

dedylja [7]

Answer:

ΔQ = 4930.37 BTu

Explanation:

given data

height h = 8ft

Δt = 8 hours

length L = 24 feet

R value = 16.2 hr⋅°F⋅ft² /Btu

inside temperature t1 = 68°F

outside temperature t2 = 16°F

to find out

number of Btu conducted

solution

we get here number of Btu conducted by this expression that s

$\frac{\Delta Q}{\Delta t} =\frac{-A}{R} (t2 -t1)$ ......................1

here A is area that is = h × L = 8 × 24 = 1492 ft²

put here value we get

$\frac{\Delta Q}{8} =\frac{-192}{16.2} (16-68)$

solve it we get

ΔQ = 4930.37 BTu

7 0

3 years ago

A point in the x-y plane is represented by its x-coordinate and y-coordinate. Design the class Point that can store and process

Black_prince [1.1K]

Answer:

#include <iostream>

#include <iomanip>

using namespace std;

class pointType

{

public:

pointType()

{

x=0;

y=0;

}

pointType::pointType(double x,double y)

{

this->x = x;

this->y = y;

}

void pointType::setPoint(double x,double y)

{

this->x=x;

this->y=y;

}

void pointType::print()

{

cout<<"("<<x<<","<<y<<")\n";

}

double pointType::getX()

{return x;

}

double pointType::getY()

{return y;

}

private:

double x,y;

};

int main()

{

pointType p2;

double x,y;

cout<<"Enter an x Coordinate for point ";

cin>>x;

cout<<"Enter an y Coordinate for point ";

cin>>y;

p2.setPoint(x,y);

p2.print();

system("pause");

return 0;

}

5 0

2 years ago