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3 years ago

Explain and show work:

Engineering

1 answer:

belka [17]3 years ago

3 0

Answer:

Explanation:

The sum of the digits of the number is ...

(4+1+3)+(4+6+5)+(7+8+9) = 8+15+24 = 47

The sum of those digits is 4+7=11, and those digits sum to 1+1 = 2.

That is, the value of the number mod 9 (or 3) is 2.

The ones digit is odd, so the value of the number mod 2 is 1.

This combination of modulo values tells you the mod 6 result is 5.

_____

<em>Additional comment</em>

We can look at the (mod2, mod3) values of the numbers 0 to 5:

0 ⇒ (0, 0)

1 ⇒ (1, 1)

2 ⇒ (0, 2)

3 ⇒ (1, 0)

4 ⇒ (0, 1)

5 ⇒ (1, 2) . . . . the mod {2, 3} results we have for the number of interest.

This process of adding up the digits repeatedly is referred to as "casting out 9s." The result of it is the modulo 9 value of the number (with 0 mapped to 9). Checking the mod 9 result of arithmetic operations is one quick way to spot certain kinds of errors. It can also be used as part of a divisibility test for 3 or 9.

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How can we prevent chemical hazards in labotary

Maurinko [17]

Answer:

We can prevent it by:

a) By wearing GOOGLES.

b) By wearing our Lab coat.

c) Fire extinguisher should always be present in the lab.

d) Hand Gloves must be worn.

e) No playing in the lab.

f) No touching of things/equipment's e.g bottles, in the lab.

g) No eating/snacking in the lab.

h) Always pay attention, no gisting.

i) Adult/qualified person must be present in the lab with pupils/students.

Explanation:

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4 0

3 years ago

At a lake location, the soil surface is 6 m under the water surface. Samples from the soil deposit underlying the lake indicate

777dan777 [17]

Answer:

The effective stress is 225.6 kN/m²

The total vertical stress is 338.4 kN/m²

Explanation:

Given;

depth of soil surface = 6m

weight unit of the soil samples = 18.8 kN/m³

Stress = force/area

Stress = unit weight x depth

Effective stress at a depth of 12 m below the soil surface:

= (18.8 kN/m³) x (12 m)

= 225.6 kN/m²

Effective stress = 225.6 kN/m²

Total vertical stress at a depth of 12 m below the soil surface:

= (18.8 kN/m³) x (6 m) + (18.8 kN/m³) x (12 m)

= 112.8 kN/m² + 225.6 kN/m²

= 338.4 kN/m²

Total vertical stress = 338.4 kN/m²

6 0

3 years ago

A drum contains 3 black balls, 5 red balls and 6 green balls. If 4 balls are selected at random what is the probability that the

posledela

Answer:

Probability of no red ball from 4 balls = 0.1258 (Approx.)

Explanation:

Given:

Number of black ball = 3

Number of red ball = 5

Number of green ball = 6

Find:

Probability of no red ball from 4 balls

Computation:

Probability of no red ball from 4 balls = 9c4 / 14c4

Probability of no red ball from 4 balls = 126 / 1001

Probability of no red ball from 4 balls = 0.12587

Probability of no red ball from 4 balls = 0.1258 (Approx.)

6 0

3 years ago

A pump of a water distribution system at 25°C is powered by a 15 kW electric motor whose efficiency is 90 percent. The water flo

IRISSAK [1]

The friction loss in the system is 3.480 kilowatts.

<h2>Procedure - Friction loss through a pump</h2><h2 /><h3>Pump model</h3><h3 />

Let suppose that the pump within a distribution system is an open system at steady state, whose mass and energy balances are shown below:

<h3>Mass balance</h3>

$\dot m_{in}-\dot m_{out} = 0$ (1)

$\dot m_{in} = \frac{\dot V_{in}}{\nu_{in}}$ (2)

$\dot m_{out} = \frac{\dot V_{out}}{\nu_{out}}$ (3)

<h3>Energy balance</h3>

$\eta \cdot \dot W_{el} + \dot m_{in}\cdot (h_{in}-h_{out}) - \dot W_{f} = 0$ (4)

Where:

$\dot m_{in}$ - Inlet mass flow, in kilograms per second.
$\dot m_{out}$ - Outlet mass flow, in kilograms per second.
$\dot V_{in}$ - Inlet volume flow, in cubic meters per second.
$\dot V_{out}$ - Outlet volume flow, in cubic meters per second.
$\nu_{in}$ - Inlet specific volume, in cubic meters per kilogram.
$\nu_{out}$ - Outlet specific volume, in cubic meters per kilogram.
$\eta$ - Pump efficiency, no unit.
$\dot W_{el}$ - Electric motor power, in kilowatts.
$h_{in}$ - Inlet specific enthalpy, in kilojoules per kilogram.
$h_{out}$ - Outlet specific enthalpy, in kilojoules per kilogram.
$\dot W$ - Work losses due to friction, in kilowatts.

<h3>Data from steam tables</h3>

From steam tables we get the following water properties at inlet and outlet:

Inlet

$p = 100\,kPa$ , $T = 25\,^{\circ}C$ , $\nu = 0.001003\,\frac{kJ}{kg}$ , $h = 104.927\,\frac{kJ}{kg}$ , Subcooled liquid

Outlet

$p = 300\,kPa$ , $T = 25\,^{\circ}C$ , $\nu = 0.001003\,\frac{kJ}{kg}$ , $h = 105.128\,\frac{kJ}{kg}$ , Subcooled liquid

<h3>Calculation of the friction loss in the system</h3>

If we know that $\dot V_{in} = 0.05\,\frac{m^{3}}{s}$ , $\nu_{in} = 0.001003\,\frac{m^{3}}{kg}$ , $h_{in} = 104.927\,\frac{kJ}{kg}$ , $h_{out} = 105.128\,\frac{kJ}{kg}$ , $\eta = 0.90$ and $\dot W_{el} = 15\,kW$ , then the friction loss in the system is:

$\dot W_{f} = \frac{\dot V_{in}}{\nu_{in}}\cdot (h_{in} - h_{out}) + \eta \cdot \dot W_{el}$

$\dot W_{f} = \left(\frac{0.05\,\frac{m^{3}}{s} }{0.001003\,\frac{m^{3}}{kg} } \right)\cdot \left(104.927\,\frac{kJ}{kg}-105.128\,\frac{kJ}{kg}\right) + (0.90)\cdot (15\,kW)$

$\dot W_{f} = 3.480\,kW$

The friction loss in the system is 3.480 kilowatts. $\blacksquare$

To learn more on pumps, we kindly invite to check this verified question: brainly.com/question/544887

6 0

2 years ago

The thrust angle is checked by referencing

anygoal [31]

In Engineering, the thrust angle is checked by referencing: C. vehicle centerline.

<h3>What is a thrust angle?</h3>

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Read more on thrust angle here: brainly.com/question/13000914

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5 0

2 years ago