3 years ago

What's the best way to plan an organized

Engineering

1 answer:

Alekssandra [29.7K]3 years ago

5 0

Answer:

Get ready and comfortable.

List all of the tasks you need to accomplish over the next week. .

Next schedule everything.

Get a planner/calender.

Cut those tasks that do not fit into your

You might be interested in

A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu

krek1111 [17]

Answer:

a) 42.08 ft/sec

b) 3366.33 ft³/sec

c) 0.235

d) 18.225 ft

e) 3.80 ft

Explanation:

Given:

b = 80ft

y1 = 1 ft

y2 = 10ft

a) Let's take the formula:

$\frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1}$

$10*2 = \sqrt{1 + 8f^2 - 1$

1 + 8f² = (20+1)²

= 8f² = 440

f² = 55

f = 7.416

For velocity of the faster moving flow, we have :

$\frac{V_1}{\sqrt{g*y_1}} = 7.416$

$V_1 = 7.416 *\sqrt{32.2*1}$

V1 = 42.08 ft/sec

b) the flow rate will be calculated as

Q = VA

VA = V1 * b *y1

= 42.08 * 80 * 1

= 3366.66 ft³/sec

c) The Froude number of the sub-critical flow.

V2.A2 = 3366.66

Where A2 = 80ft * 10ft

Solving for V2, we have:

$V_2 = \frac{3666.66}{80*10}$

= 4.208 ft/sec

Froude number, F2 =

$\frac{V_2}{g*y_2} = \frac{4.208}{32.2*10}$

F2 = 0.235

d) $El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}$

$El = \frac{(10-1)^3}{4*1*10}$

$= \frac{9^3}{40}$

= 18.225ft

e) for critical depth, we use :

$y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3$

= 3.80 ft

7 0

3 years ago

Read 2 more answers

Does anybody want 20 points? They're free get 'em while ya can...

DedPeter [7]

Answer:

Explanation:

4 0

2 years ago

Read 2 more answers

A) Total Resistance<br> b) Total Current<br> c) Current and Voltage through each resistor

Varvara68 [4.7K]

C my friend 20 characters suck

3 0

3 years ago

it creates parts from thin plastic sheets as opposed to plastic pellets. is it 1. Pickling 2. Thermoforming 3. Extrusion

iVinArrow [24]

Answer:

thermoforming

Explanation:

please mark this answer as the brainlest

4 0

3 years ago

Read 2 more answers

water at 20c discharges from a stroke tank through a 150m length of horizontal pipe. The pipe is smooth and has a diameter of 75

inessss [21]

Answer:

The dept of the water needed is 26.11m

Explanation:

Given the following parameters:

Length of the pipe = $150m$

Diameter of the pipe = $75mm = 0.75m$

Volumetric flow rate $1m^{3}/s$

Knowing that:

Volumetric flowrate $= area (A) * Velocity(V)$

==> $0.1 = A*V$ *Knowing that $V = \sqrt{2gh}$

==> $0.1 = \frac{\pi }{4}* (0.075)^{2} * \sqrt{2gh}$

==> $h = 26.11m$

Hence, the dept of water needed to produce a volumetric flow rate of $1m^{3}/s$ is 26.11m.

7 0

3 years ago

What's the best way to plan an organized​

What's the best way to plan an organized