The interactive activities in this course
A. will be tested in lesson quizzes but not the final exam
B. will be tested in the final exam but not the lesson quizzes
C. will be scored by your instructor
D. will not be score

Answer:

LCDs consume less power than CRTs

Explanation:

The sentence that best compares LCDs with CRTs is that "LCDs consume less power than CRTs".

Recently, in many homes and offices, cathode ray tube (CRT) monitors are being replaced by liquid crystal display (LCD) monitors. There are some benefits that LCD panels offer over CRT monitors. Some of them are:

LCD monitors don't usually take up much space. They take less space than CRT monitors. Also, LCDs consume less power than CRT monitors. Which means that CRTs consume more power than LCDs. LCD monitors gives a much brighter image than CRT monitors .

Given :

Capacitor , C = 55 μF .

Energy is given by :

[tex]\omega(t)=10cos^2 (377t)\ J[/tex] .

To Find :

The current through the capacitor.

Solution :

Energy in capacitor is given by :

[tex]\omega=\dfrac{Cv^2}{2}\\\\v=\sqrt{\dfrac{2\omega}{C}}\\\\v=\sqrt{\dfrac{2\times 10cos^2 (377t)}{55\times 10^{-6}}}\\\\v=cos(337t)\sqrt{\dfrac{2\times 10}{55\times 10^{-6}}}\\\\v=603.02\ cos( 337t)[/tex]

Now , current i is given by :

[tex]i=C\dfrac{dv}{dt}\\\\i=C\dfrac{d[603.02cos(337t)]}{dt}\\\\i=-55\times 10^{-6}\times 603.03\times 337\times sin(337t)\\\\i=-11.18\ sin(337t)[/tex]

( differentiation of cos x is - sin x )

Therefore , the current through the capacitor is -11.18 sin ( 377t).

Hence , this is the required solution .

Answer: C. Biotechnology system

Explanation:

Answer:

3150 L/min

Explanation:

The volume of the room is the product of the height of the room and its area. It is given as:

Volume (V) = height (h) × Area (A)

V = h × A

Height = 2.7 m, Area = 200 m². Hence:

V = h × A = 2.7 × 200 = 540 m³

The flow capacity ([tex]\dot {V}[/tex]) is given by

[tex]\dot {V}=ACH*Volume\ of\ room[/tex]

ACH = air changes per hour = 0.35

[tex]\dot {V}=ACH*Volume\ of\ room\\\\\dot {V}=0.35*540\\\\\dot {V}=189\ m^3/h[/tex]

But 1 m³ = 1000 L, 1 hr = 60 min

[tex]\dot{V}=189\ m^3/h=\frac{189\ m^3*\frac{1000\ L}{1\ m^3} }{1\ h*\frac{60\ min}{1\ h} } =3150\ L/min\\\\\dot{V}=3150 \ L/min[/tex]

Answer:

a. will be damaged if energized

Explanation:

A hermetic compressor needs a coolant to prevent overheating of the motor. The cooling of the compressor depends on the flow of the refrigerant through the compressor hence when under a deep vacuum,  the motor winding of a hermetic compressor can overheat.  

A crankcase heater is used to prevent the mixing of refrigerant and oil at low temperatures. The mixing of refrigerant and oil can cause oil foaming in the vacuum. When under a deep vacuum,  the motor winding of a hermetic compressor can overheat if the crankcase heater is energized.

Answer:

construction workers

Explanation:

A construction worker is usually someone with the technical skills and abilities needed to manually construct physical infrastructures.

Since a walkway is an infrastructure, hiring skilled construction workers should perform the project of building the series of walkways within the large garden apartment.

Answer:

voltage = -0.01116V

power = -0.0249W

Explanation:

The voltage v(t) across an inductor is given by;

v(t) = L[tex]\frac{di(t)}{dt}[/tex]             -----------(i)

Where;

L = inductance of the inductor

i(t) = current through the inductor at a given time

t = time for the flow of current

From the question:

i(t) = [tex]10e^{-t/2}[/tex]A

L = 10mH = 10 x 10⁻³H

Substitute these values into equation (i) as follows;

v(t) = [tex](10*10^{-3})\frac{d(10e^{-t/2})}{dt}[/tex]

Solve the differential

v(t) = [tex](10*10^{-3})\frac{-1*10}{2} (e^{-t/2})[/tex]

v(t) = -0.05 [tex]e^{-t/2}[/tex]

At t = 8s

v(t) = v(8) = -0.05 [tex]e^{-8/2}[/tex]

v(t) = v(8) = -0.05 [tex]e^{-4}[/tex]

v(t) = -0.05 x 0.223

v(t) = -0.01116V

(b) To get the power, we use the following relation:

p(t) = i(t) x v(t)

Power at t = 8

p(8) = i(8) x v(8)

i(8) = i(t = 8) = [tex]10e^{-8/2}[/tex]

i(8) = [tex]10e^{-4}[/tex]

i(8) = 10 x 0.223

i(8) = 2.23

Therefore,

p(8) = 2.23 x -0.01116

p(8) = -0.0249W

Answer:

The voltage is - 0.9158 mV

The power is - 0.1677 mW

Explanation:

Given;

current through the inductor, i(t)  = [tex]10e^{-t/2}[/tex] -----equation (1)

inductance, L = 10 mH

given time, t  = 8 s

The voltage across the inductor is given by;

[tex]V_L = L\frac{di}{dt} \\\\V_L = (10 *10^{-3})\frac{d}{dt} (10e^{-t/2})\\\\V_L = (10 *10^{-3})\frac{10}{-2}(e^{-t/2})\\\\ V_L = -0.05e^{-t/2} \ ----equation (2)[/tex]

when t = 8 s, the voltage will be ;

[tex]V_L = -0.05 e^{-t/2}\\\\V_L = -0.05 e^{-8/2}\\\\V_L = -0.05 e^{-4}\\\\V_L = -9.158 *10^{-4} \ V\\\\V_L = -0.9158 \ mV[/tex]

Power is given by;

P = I V

When t = 8, the current "I" is given by;

[tex]i(t) = 10e^{-t/2}\\\\i(8) = 10e^{-8/2}\\\\I = 10e^{-4}\\\\I = 0.1832 \ A[/tex]

P = 0.1832 x (-9.158 x 10⁻⁴)

P = -1.677 x 10⁻⁴ W

P = -0.1677 mW

Answer:

The world's oldest dress called the Tarkhan Dress is at 5,100 to 5,500 years of age.

Does that help? Or do you need something else? I can change my answer if this is not what you need! :D

Explanation:

The ideal mechanical advantage, actual mechanical advantage and efficiency of the wheel are respectively; 4, 3 and 75%

We are given;

Radius of crank handle; r_c = 0.4 m

Radius of axle; r_a = 0.1 m

Load = 60 N

Effort = 20 N

A) Formula for ideal mechanical advantage is;

IMA = 2πr_c/2πr_a

IMA = 0.4/0.1

IMA = 4

B) Actual mechanical advantage is;

AMA = 60/20

AMA = 3

C) Efficiency is gotten from the formula;

Efficiency = AMA/IMA * 100%

Efficiency = (3/4) * 100%

Efficiency = 75%

Read more about Mechanical Advantage at; https://brainly.com/question/18345299

Answer:

A,C and E

Explanation:

It should be understood that a function generator can be explained as a piece of electronic test equipment or software which is used in generating different types of electrical waveforms over a wide range of frequencies. It should be noted that some of the examples of waveforms produced by the function generator are the sine wave, square wave, triangular wave and sawtooth shapes.

Therefore, the reason why of picking the options as highlighted above.

✅C) Avoid having to use a net ✅

IamSugarBee

Answer:

  5 rpm

Explanation:

Each turn of the driver gear moves 10 teeth past the point of engagement with the driven gear. That is half of the 20 teeth on the driven gear, so it makes 1/2 turn for each turn of the driver.

In 1 minute, the driver makes 10 turns, so the driven gear makes 5 turns. Its speed is 5 rpm.

Answer:

d. To Murrow, Hitler was a threat to all of civilization

Explanation:

For Murrow, Hitler's rise was a serious problem and a major threat to the entire civilization. For this reason, he believed that covering news about Hitler's advance and the battles of Nazi Germany was essential, even if the American population did not see Hitler as a threat and the European population, believed that these reprotations were only a way to denigrate the image of the continent to the world.

For Murrow reporting on Hitler's actions was as important as reporting on natural disasters.

Answer:

The answer is below

Explanation:

Given:

kg (glass) = 0.78 W/m⋅K, ka (air) = 0.026 W/m⋅K, h1 = 10 W/m2⋅K, h2 = 25 W/m2⋅K, glass length = Lg = 3 mm = 0.003 m, Lo = 3 mm = 0.003 m, La = 12 mm = 0.012 Height = 1.5 m, width = 2.4 m, room temperature = T = 20°C. Therefore:

Total resistance per unit area is given as:

[tex]R"=\frac{1}{h_1}+\frac{L_g}{k_g}+\frac{L_a}{k_a} +\frac{L_o}{k_g}+\frac{1}{h_2} \\\\R"=\frac{1}{10}+\frac{0.003}{0.78}+\frac{0.012}{0.026} +\frac{0.003}{0.78}+\frac{1}{25} \\\\R"=0.1+0.00385+0.46154+0.00385+0.04\\\\R"=0.60924\ K.m^2/W[/tex]

Area = A = height * width = 1.5 m × 2.4 m = 3.6 m²

The change in temperature = ΔT = 20 °C - (-5 °C) = 25 °C

The rate of heat loss is given as:

[tex]\dot {Q}=A*\frac{\Delta T}{R"}= 3.6*\frac{25}{0.60924}\\ \\\dot {Q}=147.73\ W[/tex]

The inner surface temperature (Ti) is given as:

[tex]T_i=T-\frac{\dot {Q}}{A} *\frac{1}{h_1}\\ \\T_i=20-\frac{147.73}{3.6}*\frac{1}{10}=15.9\ ^oC[/tex]

Explanation:

It gives her a sense of purpose.

Answer:

The hot resistance of a light bulb is 205.714 ohms.

Explanation:

Hot resistance can be found by using Ohm's Law and definition of power, whose expression as a function of power and voltage is:

[tex]R = \frac{V^{2}}{\dot W}[/tex]

Where:

[tex]\dot W[/tex] - Power, measured in Watts.

[tex]V[/tex] - Voltage, measured in Volts.

[tex]R[/tex] - Resistance, measured in Ohms.

Given that [tex]V = 120\,V[/tex] and [tex]\dot W = 70\,W[/tex], we find that hot resistance is:

[tex]R = \frac{(120\,V)^{2}}{70\,W}[/tex]

[tex]R = 205.714\,\Omega[/tex]

The hot resistance of a light bulb is 205.714 ohms.

Answer:

a. From Y to X

b. q = 15 C

Explanation:

a.

When current is denoted by double subscript, it is interpreted as traveling from 1st subscript to the second, when the value of current is positive. On the other hand, if the value of current is negative, then it means that the current is traveling from 2nd subscript to the 1st subscript. Since, the value of current is negative in the given question, therefore, it means that the current is traveling from 2nd subscript to the 1st subscript. Hence the direction of current or the flow of electrons is:

From Y to X

b.

Using the following formula of current:

I = q/t

where,

I = Current (Absolute Value) = 3  A

q = amount of charge = ?

t = time taken = 5 s

Therefore,

3 A = q/5 s

q = (3 A)(5 s)

q = 15 C

Answer:

  Vab ≈ 3.426 V

Explanation:

First of all, it is convenient to find the equivalent parallel resistance of R5 and R6. That will be ...

  R56 = (R5)(R6)/(R5 +R6) = (1000)(1500)/(1000 +1500) = 600

Then we can call V1 the voltage at the top of R2. The voltage at Va is a divider from V1:

  Va = V1·(R4/(R3+R4)) = V1(560/1030) ≈ 0.543689V1

The voltage at Vb is also a divider from V1:

  Vb = V1·(R7+R8)/(R2 +R56 +R7 +R8) = V1(780/1710) ≈ 0.456140V1

The parallel branches containing Va and Vb have an effective resistance of ...

  (1030)(1710)/(1030+1710) = 642.81

That forms a divider with R1 to give V1:

  V1 = (100 V)642.81/(1000 +642.81) ≈ 39.1287 V

The difference Va-Vb is ...

  Vab = (39.1287 V)(0.543689 -0.456140) ≈ 3.426 V

_____

We have done this using parallel resistance and voltage divider calculations. You can also do it using node voltage equations. Using the same definition for V1 as above, we have ...

  (Vs -V1)/R1 +(Vb -V1)/(R56+R2) +(Va-V1)/R3 = 0

  (V1 -Vb)/(R56 +R2) -Vb/(R7+R8) = 0

  (V1 -Va)/R3 -Va/R4 = 0

The solution of interest is the value of Vab, shown in the attachment. It computes as 154200/45013 V ≈ 3.42568 V.

Answer:

k = 0.1118 per min

Explanation:

Assume;

Initial number of bacteria = N0

Number of bacteria IN 'T' time = Nt

So,

[tex]Nt=N0e^{-kt}\\\\in\ 6.2 min\\\\\\frac{N0}{2}= N0e^{-k(6.2)}\\\\ln\frac{1}{2} = -k[6.2][/tex]

k = 0.1118 per min

Answer:

bsbsdbsd

Explanation:

Marking 'N or Y' to indicate wrong or right units to your questions  :

a) N  A typical size banana gives you about 2.5 watts of energy

b) Y it takes 10kW to keep 10W light bulb on for one hour

c) Y  it takes 100 HP to lift a 1000 kg weight from ground to the 4th floor

The energy gotten from the consumption of banana will have an s.i unit of joules and not watts.  While the power consumed by an electric bulb will be measured in watts which is the s.i unit of power.

The power required to lift weight/mass ( Kg ) from the ground has an s.i unit of HP ( 1 HP = 746 watts ).

Hence we can conclude that the answers to your questions are : N , Y , Y

Learn more about s.i units : https://brainly.com/question/760437

Answer:

A. ) 591.7 v

B.) 991.7v

C.) 59.7%

D.) 47.9 Kw

E.) 247925 W

F.) 59.7 %

Explanation:

Given that a simple power system consists of a dc generator connected to a load center via a transmission line. The load power is 100 kW. The transmission line is 100 km copper wire of 3 cm diameter. If the voltage at the load side is 400 V,

Let first calculate the resistance in the wire.

The resistivity (rho) of a copper wire is 1.673×10^-8 ohm metres

Resistance R =( L× rho)/A

Where Area = πr^2 = π × 0.015^2

Area = 0.00071 m^2

R = (100000 × 1.673×10^-8) / 0.00071

Resistance in wire = 2.367 ohms

Then let calculate the resistance in the load.

Also, since Power P = V^2 /R

Make R the subject of formula

R = V^2/ P

R = 400^2/100000

Resistance in load = 1.6 Ohms

Current l = V / R

I = 400/1.6 = 250 Ampere

a.) Voltage drop across the line V line will be achieved by using Ohms law.

V = I R

V = 250 × 2.367

V = 591.7 v

B.) Voltage at the source side Vsource will be

V = V line + V load

V = 400 + 591.7

V = 991.7 v

C.) Percentage of the voltage drop Vline /Vsource

591.7/991.7 × 100 = 59.7%

D.) Line losses

P = I V

P = 250 × 591.7

P = 147925 W

Power loss = 147925 - 100000

Power loss = 47,925 W

Power loss = 47.9 Kw

E.) Power delivered by the source

P = IV

P = 250 × 991.7 = 247925 W

F.) System efficiency

Efficiency = power line / power source × 100

Efficiency = 147925 / 247925 × 100

Efficiency = 59.7 %

In this exercise we have to use the circuit knowledge of an electrical system and calculate the characteristics so we have to:

A. ) 591.7 v

B.) 991.7v

C.) 59.7%

D.) 47.9 Kw

E.) 247925 W

F.) 59.7 %

Organizing the information given in the statement we have that:

Calculating the resistivity we find that:

[tex]R =( L* \rho)/A\\A= \pi r^2 = 0.00071 m^2\\R = (100000 * 1.673*10^{-8}) / 0.00071\\R = 2.367 ohms[/tex]

Then it becomes simpler to calculate the power and current, we have:

A)  With the above information, we can calculate the voltage as:

[tex]V = I R\\V = 250 * 2.367\\V = 591.7 v[/tex]

B) Now calculating the source voltage, we find that:

[tex]V = V line + V load\\V = 400 + 591.7\\V = 991.7 v[/tex]

C.) The percentage will be calculated as the division of the two previous values, like:

[tex]591.7/991.7 * 100 = 59.7\%[/tex]

D.) Like any imperfect circuit, losses occur, so the loss will be calculated as:

[tex]Power loss = 147925 - 100000\\Power loss = 47,925 W\\Power loss = 47.9 Kw[/tex]

E.) Power delivered by the source, can be:

[tex]P = IV\\P = 250 * 991.7 = 247925 W[/tex]

F.) System efficiency, will be:

[tex]Efficiency = power line / power source *100\\Efficiency = 147925 / 247925 * 100\\Efficiency = 59.7 \%[/tex]

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Answer:

Median

Explanation:

Since the engineer is interested in the relative strength of machine parts made of two different metals. And after conducting her measurements, she learned that a small number of the plastic parts may have had cracks in them before testing, which would have caused them to break at ridiculously low force levels. Whereas, there was no possibility of cracked aluminum parts.

This small crack in the plastic part will affect the mean (average) value if it is used as a central tendency for comparison. This same thing will affect the mode and the range of the value of strength.

The best measure of central of tendency to use is median since the affected cracked part in the plastic is minimal.

Answer:

126984 cycles

Explanation:

Given data :

Sut = 530 MPa

f = 0.9

fully corrected Se = 210 MPa

using Miner's method attached below is the detailed solution of the given problem

when loaded with ± 225 MPa the number of cycles before it fails will be

≈  126984

Answer:

The python code for the given problem is written in explanation section.

Explanation:

The code for the given situation, in Python, can be written as follows:

x = int(input("Enter a positive number for dividend: "))

y = int(input("Enter a positive number for divisor: "))

quotient = x//y

remainder = x%y

print("The Quotient of the given calculation is:",quotient)

print("The Remainder of the given calculation is:",remainder)

Answer:

42.85 m/min

Explanation:

To save cost, the tools should not be changed frequently if the tool cost or tool change time is high

[tex]The\ operator\ and\ machine\ tool\ rate =C_o=\$36/hr=\frac{\$36}{1\ hr*\frac{60\ min}{1\ hr} } =\$0.6/min[/tex]

The tooling cost per cutting edge = [tex]C_t[/tex] = $4.25

n = 0.13, C=75 (m/min), tool change time = [tex]t_t=4\ min[/tex]

Therefore the cutting speed for minimum cost is given as:

[tex]v_{max}=\frac{C}{[\frac{C_o}{(\frac{1}{n}-1 )(C_o*t_t+c_t)} ]^n} \\\\Substituting:\\\\v_{max}=75{[\frac{0.6}{(\frac{1}{0.13}-1 )(0.6*4+4.25)} ]^{0.13}} \\\\v_{max}=42.85\ m/min[/tex]

Answer:

Efficiency of low pressure Turbine ( [tex]n_{lp}[/tex] ) = 82.9%

Square meters ( Area of collectors ) = 9705.2 m^2

Explanation:

Given data :

T1 = 400⁰c

p1 = 5 Mpa

p2 = 600 kpa

p3 = 600 kpa

p4 = 10 kpa

high pressure turbine efficiency = 85%

solar radiation = 900 w/m^2

Csp efficiency = 40%

A) calculate the efficiency of a low-pressure turbine

E ) Determine the area of the collectors

attached below is a detailed solution of the given problem

Answer:

Centimeters

Explanation: