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2 years ago

Chemical engineers determine how to transport chemicals. O True False

Engineering

2 answers:

Anna35 [415]2 years ago

6 0

Answer:

Chemical Engineers determine how to transport chemicals:

TRUE

Alecsey [184]2 years ago

6 0

Answer:

Explanation:

Trues

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A garden hose fills a 2-gallon bucket in 5 seconds. The number of gallons, g (y), is proportional to the number of seconds, t (x

stepladder [879]

Answer:

0.4 gallons per second

Explanation:

A function shows the relationship between an independent variable and a dependent variable.

The independent variable (x values) are input variables i.e. they don't depend on other variables while the dependent variable (y values) are output variables i.e. they depend on other variables.

The rate of change or slope or constant of proportionality is the ratio of the dependent variable (y value) to the independent variable (x value).

Given that the garden hose fills a 2-gallon bucket in 5 seconds. The dependent variable = g = number of gallons, the independent variable = t = number of seconds.

Constant of proportionality = g / t = 2 / 5 = 0.4 gallons per second

5 0

3 years ago

A particle moves along a circular path of radius 300 mm. If its angular velocity is θ = (2t) rad/s, where t is in seconds, deter

uysha [10]

Answer:

4.83m/ $s^{2}$

Explanation:

For a particle moving in a circular path the resultant acceleration at any point is the vector sum of radial and the tangential acceleration

Radial acceleration is given by $a_{radial}=w^{2}$ r

Applying values we get $a_{radial}=(2t)^{2}$ X0.3m

Thus $a_{radial}=1.2t^{2}$

At time = 2seconds $a_{radial}= 4.8m/s^{2}$

The tangential acceleration is given by $a_{tangential} =\frac{dV}{dt}=\frac{d(wr)}{dt}$

$a_{tangential}=\frac{d(2tr)}{dt}$

$a_{tangential}= 2r$

$a_{tangential}=0.6m/s^{2}$

Thus the resultant acceleration is given by

$a_{res} =\sqrt{a_{rad}^{2}+a_{tangential}^{2}}$

$a_{res} =\sqrt{4.8^{2}+0.6^{2} } =4.83m/s^{2}$

8 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 cal

Sphinxa [80]

Answer:

a) The velocity is 42.0833 ft/s

b) The flow rate is 3366.664 ft³/s

c) The Froude number is 0.2345

d) The flow energy dissipated (expressed as percentage of the energy prior to the jump) is 18.225 ft

e) The critical depth is 3.8030 ft

Explanation:

Given data:

80 ft wide channel, L

1 ft and 10 ft water depths, d₁ and d₂

Questions: a) Velocity of the faster moving flow, v = ?

b) The flow rate (discharge), q = ?

c) The Froude number, F = ?

d) The flow energy dissipated, E = ?

e) The critical depth, dc = ?

a) For the velocity:

$\frac{d_{2} }{d_{1} } =\frac{1}{2} (\sqrt{1+8F^{2} } -1)$

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

Solving for F:

F = 7.4162

$v=F\sqrt{gd_{1} }$

Here, g = gravity = 32.2 ft/s²

$v=7.4162*\sqrt{32.2*1} =42.0833ft/s$

b) The flow rate:

$q=v*L*d_{1} =42.0833*80*1=3366.664ft^{3} /s$

c) The Froude number:

$v_{2} =\frac{q}{L*d_{2} } =\frac{3366.664}{80*10} =4.2083ft/s$

$F=\frac{v_{2}}{\sqrt{gd_{2} } } =\frac{4.2083}{\sqrt{32.2*10} } =0.2345$

d) The flow energy dissipated:

$E=\frac{(d_{2}-d_{1})^{3} }{4d_{1}d_{2}} =\frac{(10-1)^{3} }{4*1*10} =18.225ft$

e) The critical depth:

$d_{c} =(\frac{(\frac{q}{L})^{2} }{g} )^{1/3} =(\frac{(\frac{3366.664}{80})^{2} }{32.2} )^{1/3} =3.8030ft$

4 0

4 years ago

An incoming signal is at a frequency of 500kHz. This signal needs to be acquired and all other signals attenuated. Design a pass

Finger [1]

Answer:

C_h = 0.166 nF

C_L = 0.153 nF

Explanation:

Given:

- Ideal frequency f_o = 500 KHz

- Bandwidth of frequency BW = 40 KHz

- The resistance identical to both low and high pass filter = 2 Kohms

Find:

Design a passive band-pass filter to do this by cascading a low and high pass filter.

Solution:

- First determine the cut-off frequencies f_c for each filter:

f_c,L for High pass filter:

f_c,L = f_o - BW/2 = 500 - 40/2

f_c,L = 480 KHz

f_c,h for Low pass filter:

f_c,h = f_o + BW/2 = 500 + 40/2

f_c,h = 520 KHz

- Now use the design formula for R-C circuit for each filter:

General design formula:

f_c = 1 /2*pi*R*C_i

C,h for High pass filter:

C_h = 1 /2*pi*R*f_c,L

C_h = 1 /2*pi*2000*480,000

C_h = 0.166 nF

C,L for Low pass filter:

C_L = 1 /2*pi*R*f_c,h

C_L = 1 /2*pi*2000*520,000

C_L = 0.153 nF

7 0

3 years ago

Hellpppppppppp will give brainliest just help!!!!!!

aliya0001 [1]

Turbooooooooooi

Hope it helps Mister

3 0

3 years ago