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2 years ago

Are you?

Engineering

2 answers:

Inessa [10]2 years ago

6 0

Answer:

are you wht

didn't understand the question

AlladinOne [14]2 years ago

3 0

Answer:

Yes/No

Explanation:

NANI?!

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The purification of hydrogen gas is possible by diffusion through a thin palladium sheet. Calculate the number of kilograms of h

gtnhenbr [62]

Answer: 5.36×10-3kg/h

Where 10-3 is 10 exponential 3 or 10 raised to the power of -3.

Explanation:using the formula

M =JAt = -DAt×Dc/Dx

Where D is change in the respective variables. Insulting the values we get,

=5.1 × 10-8 × 0.13 × 3600 × 2.9 × 0.31 / 4×10-3.

=5.36×10-3kg/h

6 0

3 years ago

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

When block C is in position xC = 0.8 m, its speed is 1.5 m/s to the right. Find the velocity of block A at this instant. Note th

Amanda [17]

Answer:

The answer is "2 m/s".

Explanation:

The triangle from of the right angle:

$\to (x_c-0.8)+(1.5+y_4) +\sqrt{x_c^2 + 1.5^2}= constant$

Differentiating the above equation:

$\to V_c +V_A+ \frac{X_cV_c}{\sqrt{x_c^2 +1}}=0\\\\\to 1-V_A+ \frac{0.8 \times 1.5}{\sqrt{ 0.8^2+1.5}}=0\\\\$

$\to V_A= \frac{1.2}{\sqrt{ 0.64+1.5}}+1\\\\$

$= \frac{1.2}{ 1.46}+1\\\\= \frac{1.2+ 1.46}{ 1.46}\\\\ = \frac{2.66}{1.46}\\\\= 1.82 \ \frac{m}{s}\\\\= 2 \ \frac{m}{s}$

3 0

2 years ago

What kinds of problems or projects would a civil engineer work on?

lisov135 [29]

Answer:

simple projects bovonhztisgx

8 0

3 years ago

A microwave transmitter has an output of 0.1W at 2 GHz. Assume that this transmitter is used in a microwave communication system

Len [333]

Answer:

gain = 353.3616

P_r = 1.742*10^-8 W

Explanation:

Given:

- The output Power P_o = 0.1 W

- The diameter of the antennas d = 1.2 m

- The frequency of signal f = 2 GHz

Find:

a. What is the gain of each antenna?

b. If the receiving antenna is located 24 km from the transmitting antenna over a free space path, find the available signal power out of the receiving antenna.

Solution:

- The gain of the parabolic antenna is given by the following formula:

gain = 0.56 * 4 * pi^2 * r^2 / λ^2

Where, λ : The wavelength of signal

r: Radius of antenna = d / 2 = 1.2 / 2 = 0.6 m

- The wavelength can be determined by:

λ = c / f

λ = (3*10^8) / (2*10^9)

λ = 0.15 m

- Plug in the values in the gain formula:

gain = 0.56 * 4 * pi^2 * 0.6^2 / 0.15^2

gain = 353.3616

- The available signal power out from the receiving antenna is:

P_r = (gain^2 * λ^2 * W) / (16*pi^2 * 10^2 * 10^6)

P_r = (353.36^2 * 0.15^2 * 0.1) / (16*pi^2 * 10^2 * 10^6)

P_r = 1.742*10^-8 W

4 0

3 years ago