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Tema [17]
3 years ago
13

How do i do this? if y’all don’t mind helping lol

Engineering
1 answer:
Nookie1986 [14]3 years ago
7 0
Get a ruler and measure it and for every 1/8 of an inch you multiply by 1 to get your footage
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An insulated, vertical piston-cylinder device initially contains 10kg of water, 6kg of which is in the vapor phase. The mass of
Alexeev081 [22]

Answer:

a)120C

b)29kg

Explanation:

Hello!

To solve this exercise follow the steps below

1. we will call 1 the initial state, 2 the steam that enters and 3 the final state

2. We find the quality of the initial state, dividing the mass of steam by the total mass.

q1=\frac{6kg}{10kg} =0.6

3 Find the internal energy in the three states using thermodynamic tables

note:Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  

u1=IntEnergy(Water;x=0,6(quality);P=200kPa) =1719KJ/kg

u2=IntEnergy(Water;t=350;P=5000kPa) =2808KJ/kg

u3=IntEnergy(Water;x=1;P=200kPa) =2529KJ/kg

4. use the internal energy and pressure to find the temperature in state 3, using thermodynamic tables

T3=Temperature(Water;P=200kPa;u=u3=2529KJ/kg)=120C

5. Use the first law of thermodynamics in the system, it states that the initial energy in a system must be equal to the final

m1u1+m2u2=(m1+m2)u3

where

m1=inital mass=10kg

m2=the mass of the steam that has entered.

solve for m2

(m1)(u1-u3)=(m2)(u3)-(m2)(u2)

m2=m1\frac{u1-u3}{u3-u2} =10\frac{1719-2529}{2529-2808} =29kg

7 0
3 years ago
ILL GIVE YOU BRAINLIEST PLEASE HELP PLEASE
Lesechka [4]

Answer:

try to pop it back in good luck im scared for you

4 0
3 years ago
Doubling the diameter of a solid, cylindrical wire doubles its strength in tension.
julsineya [31]

Answer:

True ❤️

-Solid by solid can make Cylindrical wire doubles Strengths in tension

4 0
3 years ago
A fatigue test was conducted in which the mean stress was 90 MPa (13050 psi), and the stress amplitude was 190 MPa (27560 psi).
Gwar [14]

Answer:

a) 280MPa

b) -100MPa

c) -0.35

d) 380 MPa

Explanation:

GIVEN DATA:

mean stress \sigma_m = 90MPa

stress amplitude \sigma_a = 190MPa

a) \sigma_m =\frac{\sigma_max+\sigma_min}{2}

    90 =\frac{\sigma_{max}+\sigma_{min}}{2} --------------1

\sigma_a =\frac{\sigma_{max}-\sigma_{min}}{2}

   190 = \frac{\sigma_{max}-\sigma_{min}}{2} -----------2

solving 1 and 2 equation we get

\sigma_{max} = 280MPa

b) \sigma_{min} = - 100MPa

c)

stress ratio=\frac{\sigma_{min}}{\sigma_{max}}

=\frac{-100}{280} = -0.35

d)magnitude of stress range

                      =(\sigma_{max} -\sigma_{min})

                       = 280 -(-100) = 380 MPa

3 0
3 years ago
Wright Company deposits all cash receipts on the day when they are received and it makes all cash payments by check. At the clos
alina1380 [7]

Answer:

                                              Wright Company

                                             Bank Reconciliation

                                                 May 31, 2013

Credit side                                                                                   Debit side

Bank statement $26200                 |                          Book balance $27900

<em>Add;                                                    </em>

Deposit on May 31 $6400

Bank error $420

Sub-total=$33020

Deductions;                                        |                       Deduct

ions

Outstanding checks $5800              |                 Bank service charge $120

Adjusted bank balance $27220       |                  NSF check $560

                                                                             Total deduction $680

                                                      Adjusted book balance $27220

8 0
3 years ago
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