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3 years ago

What are some quality assurance systems

1 answer:

7 0

Examples of quality assurance activities include process checklists, process standards, process documentation and project audit. Examples of quality control activities include inspection, deliverable peer reviews and the software testing process. You may like to read more about the quality assurance vs quality control.

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A BS of 5.43 ft is taken on a level rod at a 120-ft distance, and a FS of 8.76 ft is taken on the rod held 1,100 feet away.(a) W

jok3333 [9.3K]

Answer:

The answer is attached below

Explanation:

3 0

3 years ago

Two substances, A and B, initially at different temperatures, come into contact and reach thermal equilibrium. The mass of subst

Kaylis [27]

Answer:

The specific heat capacity of substance A is 1.16 J/g

Explanation:

The substances A and B come to a thermal equilibrium, therefore, the heat given by the hotter substance B is absorbed by the colder substance A.

The equation becomes:

Heat release by Substance B = Heat Gained by Substance A

The heat can be calculated by the formula:

Heat = mCΔT

where,

m = mass of substance

C = specific heat capacity of substance

ΔT = difference in temperature of substance

Therefore, the equation becomes:

(mCΔT) of A = (mCΔT) of B

FOR SUBSTANCE A:

m = 6.01 g

ΔT = Final Temperature - Initial Temperature

ΔT = 46.1°C - 20°C = 26.1°C

C = ?

FOR SUBSTANCE B:

m = 25.6 g

ΔT = Initial Temperature - Final Temperature

ΔT = 52.2°C - 46.1°C = 6.1°C

C = 1.17 J/g

Therefore, eqn becomes:

(6.01 g)(C)(26.1°C) = (25.6 g)(1.17 J/g)(6.1°C)

C = (182.7072 J °C)/(156.861 g °C)

C = 1.16 J/g

5 0

3 years ago

A well insulated turbine operates at steady state. Steam enters the turbine at 4 MPa with a specific enthalpy of 3015.4 kJ/kg an

Anarel [89]

Answer:

power developed by the turbine = 6927.415 kW

Explanation:

given data

pressure = 4 MPa

specific enthalpy h1 = 3015.4 kJ/kg

velocity v1 = 10 m/s

pressure = 0.07 MPa

specific enthalpy h2 = 2431.7 kJ/kg

velocity v2 = 90 m/s

mass flow rate = 11.95 kg/s

solution

we apply here thermodynamic equation that

energy equation that is

$h1 + \frac{v1}{2} + q = h2 + \frac{v2}{2} + w$

put here value with

turbine is insulated so q = 0

so here

$3015.4 *1000 + \frac{10^2}{2} = 2431.7 * 1000 + \frac{90^2}{2} + w$

solve we get

w = 579700 J/kg = 579.7 kJ/kg

and

W = mass flow rate × w

W = 11.95 × 579.7

W = 6927.415 kW

power developed by the turbine = 6927.415 kW

7 0

3 years ago

Initially when 1000.00 mL of water at 10oC are poured into a glass cylinder, the height of the water column is 1000.00 mm. The w

Dafna11 [192]

Answer:

$\mathbf{h_2 =1021.9 \ mm}$

Explanation:

Given that :

The initial volume of water $V_1$ = 1000.00 mL = 1000000 mm³

The initial temperature of the water $T_1$ = 10° C

The height of the water column h = 1000.00 mm

The final temperature of the water $T_2$ = 70° C

The coefficient of thermal expansion for the glass is ∝ = $3.8*10^{-6 } mm/mm \ per ^oC$

The objective is to determine the the depth of the water column

In order to do that we will need to determine the volume of the water.

We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:

At temperature $T_1 = 10 ^ 0C$ the density of the water is $\rho = 999.7 \ kg/m^3$

At temperature $T_2 = 70^0 C$ the density of the water is $\rho = 977.8 \ kg/m^3$

The mass of the water is $\rho V = \rho _1 V_1 = \rho _2 V_2$

Thus; we can say $\rho _1 V_1 = \rho _2 V_2$ ;

⇒ $999.7 \ kg/m^3*1000 \ mL = 977.8 \ kg/m^3 *V_2$

$V_2 = \dfrac{999.7 \ kg/m^3*1000 \ mL}{977.8 \ kg/m^3 }$

$V_2 = 1022.40 \ mL$

$v_2 = 1022400 \ mm^3$

Thus, the volume of the water after heating to a required temperature of $70^0C$ is 1022400 mm³

However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:

$V_1 = A_1 *h_1$

The area of the water before heating is:

$A_1 = \dfrac{V_1}{h_1}$

$A_1 = \dfrac{1000000}{1000}$

$A_1 = 1000 \ mm^2$

The area of the heated water is :

$A_2 = A_1 (1 + \Delta t \alpha )^2$

$A_2 = A_1 (1 + (T_2-T_1) \alpha )^2$

$A_2 = 1000 (1 + (70-10) 3.8*10^{-6} )^2$

$A_2 = 1000.5 \ mm^2$

Finally, the depth of the heated hot water is:

$h_2 = \dfrac{V_2}{A_2}$

$h_2 = \dfrac{1022400}{1000.5}$

$\mathbf{h_2 =1021.9 \ mm}$

Hence the depth of the heated hot water is $\mathbf{h_2 =1021.9 \ mm}$

4 0

3 years ago

1. A loss of braking effectiveness due to too much heat in the brakes is called

motikmotik

Answer:

brake fade. loss of brake effectiveness due to overheating.

Explanation:

8 0

1 year ago