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2 years ago

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(a) Aluminum foil used for storing food weighs about 0.3 grams per square inch. How many atoms of aluminum are contained in one

square inch of the foil? (b) Using the densities and atomic weights given in Appendix A, calculate and compare the number of atoms per cubic centimeter in (i) lead and (ii) lithium.

1 answer:

atroni [7]2 years ago

3 0

Answer:

note:

solution is attached due to error in mathematical equation. please find the attachment

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The length of the arm is the main part of natur

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A gas has an initial volume o.25m^3, and absolute pressure 100kPa. Its initial temperature is 290k. The gas is compressed into a

dezoksy [38]

Answer:

<h2>698.3Kpa</h2>

Explanation:

Step one:

given data

V1=0.25m^3

T1=290k

P1=100kPa

V2=0.5m^2

T2=405k

P2=? final pressure

Step two:

The combined gas equation is given as

P1V1/T1=P2V2/T2

Substituting we have

(100*0.25)/290=P2*0.05/405

25/290=0.5P2/405

0.086=0.05P2/405

cross multiply

0.086*405=0.05P2

34.9=0.05P2

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P2=34.9/0.05

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5 0

2 years ago

A controller on an electronic arcade game consists of a variable resistor connected across the plates of a 0.227 μF capacitor. T

anyanavicka [17]

Answer:

R min = 28.173 ohm

R max = 1.55 × $10^{4}$ ohm

Explanation:

given data

capacitor = 0.227 μF

charged to 5.03 V

potential difference across the plates = 0.833 V

handled effectively = 11.5 μs to 6.33 ms

solution

we know that resistance range of the resistor is express as

V(t) = $V_o \times e^{t\RC}$ ...........1

so R will be

R = $\frac{t}{C\times ln(\frac{V_o}{V})}$ ....................2

put here value

so for t min 11.5 μs

R = $\frac{11.5}{0.227\times ln(\frac{5.03}{0.833})}$

R min = 28.173 ohm

and

for t max 6.33 ms

R max = $\frac{6.33}{11.5} \times 28.173$

R max = 1.55 × $10^{4}$ ohm

4 0

2 years ago

Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe

NemiM [27]

Answer:

(a) attached below

(b) $pf_{C}=0.85$ $lagging$

(c) $I_{C} =32.37 A$

(d) $X_{C} =49.37$ Ω

(e) $I_{cap} =9.72 A$ and $I_{line} =27.66 A$

Explanation:

Given data:

$P_{1}=15 kW$

$S_{2} =10 kVA$

$pf_{1} =0.6$ $lagging$

$pf_{2}=0.8$ $leading$

$V=480$ $Volts$

(a) Draw the power triangle for each load and for the combined load.

$\alpha_{1}=cos^{-1} (0.6)=53.13$ °

$\alpha_{2}=cos^{-1} (0.8)=36.86$ °

$S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA$

$Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99$ ≅ $20kVAR$

$P_{2} =S_{2}*pf_{2} =10*0.8=8 kW$

$Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99$ ≅ $-6 kVAR$

The negative sign means that the load 2 is providing reactive power rather than consuming

Then the combined load will be

$P_{c} =P_{1} +P_{2} =15+8=23 kW$

$Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR$

(b) Determine the power factor of the combined load and state whether lagging or leading.

$S_{c} =P_{c} +jQ_{c} =23+14j$

or in the polar form

$S_{c} =26.92$ °

$pf_{C}=cos(31.32) =0.85$ $lagging$

The relationship between Apparent power S and Current I is

$S=VI^{*}$

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

(c) Determine the magnitude of the line current from the source.

Current of the combined load can be found by

$I_{C} =S_{C}/\sqrt{3}*V$

$I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A$

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

$Q_{C} =3*V^2/X_{C}$

$X_{C} =3*V^2/Q_{C}$

$X_{C} =3*(480)^2/14*10^3$ Ω

(e) Compute the magnitude of the current in each capacitor and the line current from the source.

Current flowing in the capacitor is

$I_{cap} =V/X_{C} =480/49.37=9.72 A$

Line current flowing from the source is

$I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A$

8 0

3 years ago

A heat engine is coupled with a dynamometer. The length of the load arm is 900 mm. The spring balance reading is 16. Applied wei

miss Akunina [59]

Answer:

P = 80.922 KW

Explanation:

Given data;

Length of load arm is 900 mm = 0.9 m

Spring balanced read 16 N

Applied weight is 500 N

Rotational speed is 1774 rpm

we know that power is given as

$P = T\times \omega$

T Torque = (w -s) L = (500 - 16)0.9 = 435.6 Nm

$\omega$ angular speed $=\frac{2 \pi N}{60}$

Therefore Power is

$P =\frac{435.6 \time 2 \pi \times 1774}{60} = 80922.65 watt$

P = 80.922 KW

4 0

3 years ago