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3 years ago

7

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the y

-axis.
y = 5x2, y = 30x − 10x2

1 answer:

kow [346]3 years ago

3 0

Answer:

40π

Explanation:

First, find the limits (intersections).

5x² = 30x − 10x²

15x² − 30x = 0

x² − 2x = 0

x (x − 2) = 0

x = 0 or 2

Within this interval, 30x − 10x² is greater than 5x².

Dividing the volume into cylindrical shells, the volume of each shell is:

dV = 2π r h t

dV = 2π x (30x − 10x² − 5x²) dx

dV = 2π x (30x − 15x²) dx

dV = 30π (2x² − x³) dx

The total volume is the sum (integral):

V = ∫ dV

V = ∫₀² 30π (2x² − x³) dx

V = 30π ∫₀² (2x² − x³) dx

V = 30π (⅔ x³ − ¼ x⁴)|₀²

V = 30π (⅔ 8 − ¼ 16)

V = 30π (16/3 − 4)

V = 10π (16 − 12)

V = 40π

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A 75-hp motor that has an efficiency of 91.0% is worn-out and is replaced by a motor that has a high efficiency 75-hp motor that

damaskus [11]

Answer:

the reduction in the heat gain is 2.8358 kW

Given that;

Shaft outpower of a motor = 75 hp = ( 75 × 746 ) = 55950 W

Efficiency of motor = 91.0% = 0.91

High Efficiency of the motor = 95.4% = 0.954

now, we know that, efficiency of motor is defined as; = /

where is the electric input given to the motor

so

= /

we substitute

= 55950 W / 0.91

= 61483.5 W

= 61.4835 kW

now, the electric input given to the motor due to increased efficiency will be;

= /

we substitute

= 55950 W / 0.954

= 58647.79 W

= 58.6477 kW

so the reduction of the heat gain of the room due to higher efficiency will be;

Q = -

we substitute

Q = 61.4835 kW - 58.6477 kW

Q = 2.8358 kW

Therefore, the reduction in the heat gain is 2.8358 kW

Explanation: i hope this answer your question if this is wrong or correct please let me know.also no trying to be rude but can you sent me like a thanks?

8 0

2 years ago

Suppose you have a 9.00 V battery, a 2.00 μF capacitor, and a 7.40 μF capacitor. (a) Find the charge and energy stored if the ca

Andru [333]

Answer:

$Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J$

$Q=9.4*10^-6*9=8.46*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *9.4*10^-6=3.81*10^-4J$

Explanation:

<u>a)</u>

<u>Identify the unknown: </u>

The charge and energy stored if the capacitors are connected in series

<u>List the Knowns: </u>

Capacitance of the first capacitor: $C_{1}$ = 2цF = 2 x 10-6 F

Capacitance of the second capacitor $C_{2}$ = 7.4цF = 7.4 x 10-6 F

Voltage of battery: V = 9 V

<u>Set Up the Problem: </u>

Capacitance of a series combination:

$\frac{1}{C_{s} } =\frac{1}{C_{1} } +\frac{1}{C_{2} } +\frac{1}{C_{3} }$ +............

$\frac{1}{C_{s} } =\frac{1}{2} +\frac{1}{ 7.4} \\C_{s} =\frac{2*7.4}{2+7.4}=1.575 *10^-6 F\\$

Capacitance of a series combination is given by:

$C_{s}=\frac{Q}{V}$

Then the charge stored in the series combination is:

$Q=C_{s} V$

Energy stored in the series combination is:

$U_{c}=\frac{1}{2} V^{2} C_{s}$

<u>Solve the Problem: </u>

$Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J$

<u>b)</u>

<u>Identify the unknown: </u>

The charge and energy stored if the capacitors are connected in parallel

<u>Set Up the Problem: </u>

Capacitance of a parallel combination:

$C_{p} =C_{1} +C_{2} +C_{3}$

$C_{p} =2+7.4=9.4*10^-6F$

Capacitance of a parallel combination is given by

$C_{p} =\frac{Q}{V}$

Then the charge stored in the parallel combination is

$Q=C_{p} V$

Energy stored in the parallel combination is:

$U_{c}=\frac{1}{2} V^2C_{p}$

<u>Solve the Problem: </u><em> </em>

$Q=9.4*10^-6*9=8.46*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *9.4*10^-6=3.81*10^-4J$

5 0

3 years ago

Read 2 more answers

Methane and oxygen react in the presence of a catalyst to form formaldehyde. In a parallel reaction, methane is oxidized to carb

Nezavi [6.7K]

Answer:

$y_{CH_4}^2=\frac{5mol/s}{100mol/s}=0.05\\y_{O_2}^2=\frac{3mol/s}{100mol/s}=0.03\\y_{H_2O}^2=\frac{47mol/s}{100mol/s}=0.47\\y_{HCHO}^2=\frac{43mol/s}{100mol/s}=0.43\\y_{CO_2}^2=\frac{2mol/s}{100mol/s}=0.02$

Explanation:

Hello,

a. On the attached document, you can see a brief scheme of the process. Thus, to know the degrees of freedom, we state the following unknowns:

- $\xi_1$ and $\xi_2$ : extent of the reactions (2).

- $F_{O_2}^2$ , $F_{CH_4}^2$ , $F_{H_2O}^2$ , $F_{HCHO}^2$ and $F_{CO_2}^2$ : Molar flows at the second stream (5).

On the other hand, we've got the following equations:

- $F_{O_2}^2=50mol/s-\xi_1-2\xi_2$ : oxygen mole balance.

- $F_{CH_4}^2=50mol/s-\xi_1-\xi_2$ : methane mole balance.

- $F_{H_2O}^2=\xi_1+2\xi_2$ : water mole balance.

- $F_{HCHO}^2=\xi_1$ : formaldehyde mole balance.

- $F_{CO_2}^2=\xi_2$ : carbon dioxide mole balance.

Thus, the degrees of freedom are:

$DF=7unknowns-5equations=2$

It means that we need two additional equations or data to solve the problem.

b. Here, the two missing data are given. For the fractional conversion of methane, we define:

$0.900=\frac{\xi_1+\xi_2}{50mol/s}$

And for the fractional yield of formaldehyde we can set it in terms of methane as the reagents are equimolar:

$0.860=\frac{F_{HCHO}^2}{50mol/s}$

In such a way, one realizes that the output formaldehyde's molar flow is:

$F_{HCHO}^2=0.860*50mol/s=43mol/s$

Which is equal to the first reaction extent $\xi_1$ , therefore, one computes the second one from the fractional conversion of methane as:

$\xi_2=0.900*50mol/s-\xi_1\\\xi_2=0.900*50mol/s-43mol/s\\\xi_2=2mol/s$

Now, one computes the rest of the output flows via:

- $F_{O_2}^2=50mol/s-43mol/s-2*2mol/s=3mol/s$

- $F_{CH_4}^2=50mol/s-43mol/s-2mol/s=5mol/s$

- $F_{H_2O}^2=43mol/s+2*2mol/s=47mol/s$

- $F_{HCHO}^2=43mol/s$

- $F_{CO_2}^2=2mol/s$

The total output molar flow is:

$F_{O_2}+F_{CH_4}+F_{H_2O}+F_{HCHO}+F_{CO_2}=100mol/s$

Therefore the output stream composition turns out into:

$y_{CH_4}^2=\frac{5mol/s}{100mol/s}=0.05\\y_{O_2}^2=\frac{3mol/s}{100mol/s}=0.03\\y_{H_2O}^2=\frac{47mol/s}{100mol/s}=0.47\\y_{HCHO}^2=\frac{43mol/s}{100mol/s}=0.43\\y_{CO_2}^2=\frac{2mol/s}{100mol/s}=0.02$

Best regards.

7 0

3 years ago

Question 2: (a) In your own words, clearly distinguish and differentiate between Ethics in Engineering and Ethics in Computing (

zlopas [31]

Engineering ethics is not without abstraction, but in contrast with computing, it is animated by a robust and active movement concerned with the seamless identification of ethics with practice.

<h3 /><h3>What is engineering?</h3>

This is a branch of science and technology concerned with the design, building, and use of engines, machines, and structures that uses scientific principles.

Comparing ethics in engineering and ethics in computing:

Engineering ethics are a set of rules and guidelines. While computing ethics deals with procedures, values and practices.
In engineering ethics, engineers must adhere to these rules as a moral obligation to their profession While in computing ethics, the ethics govern the process of consuming computer technology.
Following these ethics for the two professions will NOT cause damage, but disobeying them causes damage.

Some practical examples in the computing field:

Avoid using the computer to harm other people such as creating a bomb or destroying other people's work.
Users also should not use a computer for stealing activities like breaking into a bank or company.
Make sure a copy of the software had been paid for by the users before it is used.

Some practical examples in the engineering field:

Integrity for oneself.
Respect for one another.
Pursuit of excellence and accountability.

Hence, Engineering ethics is the field of system of moral principles that apply to the practice of engineering and following them is important to the profession.

Read more about <em>engineering</em> here:

brainly.com/question/17169621

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7 0

2 years ago

Intravenous infusions are usually driven by gravity by hanging the bottle at a sufficient height to counteract the blood pressur

Bingel [31]

Answer:

(a) BP = 11.99 KPa

(b) h = 2 m

Explanation:

(a)

Since, the fluid pressure and blood pressure balance each other. Therefore:

BP = ρgh

where,

BP = Blood Pressure

ρ = density of fluid = 1020 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = height of fluid = 1.2 m

Therefore,

BP = (1020 kg/m³)(9.8 m/s²)(1.2 m)

<u>BP = 11995.2 Pa = 11.99 KPa</u>

(b)

Again using the equation:

P = ρgh

with data:

P = Gauge Pressure = 20 KPa = 20000 Pa

ρ = density of fluid = 1020 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = height of fluid = ?

Therefore,

20000 Pa = (1020 kg/m³)(9.8 m/s²)h

<u>h = 2 m</u>

7 0

3 years ago