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3 years ago

Which is the correct order for handwashing

Engineering

2 answers:

cupoosta [38]3 years ago

8 0

Answer:

Follow these five steps every time.

1.Wet your hands with clean, running water (warm or cold), turn off the tap, and apply soap.

2.Lather your hands by rubbing them together with the soap. Lather the backs of your hands, between your fingers, and under your nails.

3.Scrub your hands for at least 20 seconds. Need a timer? Hum the “Happy Birthday” song from beginning to end twice.

4.Rinse your hands well under clean, running water.

5.Dry your hands using a clean towel or air dry them.

Jlenok [28]3 years ago

8 0

There are 5 steps in order for you to wash your hands
1.dry them
2.wash them with soap and water or handsinitizer so it kills virus germs

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What is the purpose of the graphic language?

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Answer:

enables the representation, analysis and communication of various aspects of an information system. These aspects correspond to varying and incomplete views of information systems and the processes therein.

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3 years ago

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3 years ago

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A car is alone at a red light. When the lights turn green, it's starts at constant acceleration after traveling 100. M, it's is

Leto [7]

The acceleration of the car is 1.125 m/s².

Explanation:

Acceleration refers to the change in velocity of an object per unit time.

There are three basic formulae linking initial velocity ("u"), final velocity ("v"), distance ("s"), time ("t") and acceleration ("a"). They are-

v²=u²+2as
s =u*t+¹/₂(a*t²)
v =u+a*t

All the variable carry the meaning as explained above using formula no 1

Since the car starts from rest (initial velocity "u"=0)

Putting "u"=0 in the formula 1

we get v²=2*a*s

Thus acceleration "a" can be written as a=v²/2s

Putting the value of "v" as 15 m/s and s= 100 m (given in the question)

a=15*15 /(2*100)

a=1.125 m/s²

Thus the acceleration of the car is 1.125 m/s²

3 0

3 years ago

A rigid tank of volume of 0.06 m^3 initially contains a saturated mixture of liquid and vapor of H2O at a pressure of 15 bar and

Serga [27]

Answer:

The total mass in the tank = 0.45524 kg

The amount of heat transferred = 3426.33 kJ

Explanation:

Given that:

The volume of the tank V = 0.06 m³

The pressure of the liquid and the vapor of H2O (p) = 15 bar

The initial quality of the mixture $\mathbf{x_{initial} - 0.20}$

By applying the energy rate balance equation;

$\dfrac{dU}{dt} = Q_{CV} - m_eh_e$

where;

$m_e =- \dfrac{dm_{CV}}{dt}$

Thus, $\dfrac{dU}{dt} =Q_{CV} + \dfrac{dm_{CV}}{dt}h_e$

If we integrate both sides; we have:

$\Delta u_{CV} = Q_{CV} + h _e \int \limits ^2_1 \ dm_{CV}$

$m_2u_2 - m_1 u_1 = Q_{CV} + h_e (M_2-m_1) \ \ \ --- (1)$

We obtain the following data from the saturated water pressure tables, at p = 15 bar.

Since:

$h_e =h_g$

Then: $h_g = h_e = 2792.2 \ kJ/kg$

$v_f = 1.1539 \times 10^{-3} \ m^3 /kg$

$v_g = 0.1318 \ m^3/kg$

Hence;

$v_1 = v_f + x_{initial} ( v_g-v_f)$

$v_1 = 1.1529 \times 10^{-3} + 0.2 ( 0.1318-1.159\times 10^{-3} )$

$v_1 = 0.02728 \ m^3/kg$

Similarly; we obtained the data for $u_f \ \& \ u_g$ from water pressure tables at p = 15 bar

$u_f = 843.16 \ kJ/kg\\\\ u_g = 2594.5 \ kJ/kg$

Hence;

$u_1 = u_f + x_{initial } (u_g -u_f)$

$u_1 =843.16 + 0.2 (2594.5 -843.16)$

$u_1 = 1193.428$

However; the initial mass $m_1$ can be calculated by using the formula:

$m_1 = \dfrac{V}{v_1}$

$m_1 = \dfrac{0.06}{0.02728}$

$m_1 = 2.1994 \ kg$

From the question, given that the final quality; $x_2 = 1$

$v_2 = v_f + x_{final } (v_g - v_f)$

$v_2 = 1.1539 \times 10^{-3} + 1(0.1318 -1.1539 \times 10^{-3})$

$v_2 = 0.1318 \ m^3/kg$

Also;

$u_2 = u_f + x_{final} (u_g - u_f)$

$u_2 = 843.16 + 1 (2594.5 - 843.16)$

$u_2 = 2594.5 \ kJ/kg$

Then the final mass can be calculated by using the formula:

$m_2 = \dfrac{V}{v_2}$

$m_2 = \dfrac{0.06}{0.1318}$

$m_2 = 0.45524 \ kg$

Thus; the total mass in the tank = 0.45524 kg

FInally; from the previous equation (1) above:

$m_2u_2 - m_1 u_1 = Q_{CV} + h_e (M_2-m_1) \ \ \ --- (1)$

$Q = (m_2u_2-m_1u_1) - h_e(m_2-m_1)$

Q = [(0.45524)(2594.5) -(2.1994)(1193.428)-(2792.2)(0.45524-2.1994)]

Q = [ 1181.12018 - 2624.825543 - (2792.2)(-1.74416 )]

Q = 3426.33 kJ

Thus, the amount of heat transferred = 3426.33 kJ

7 0

3 years ago