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3 years ago

10

Given numRows and numColumns, print a list of all seats in a theater. Rows are numbered, columns lettered, as in 1A or 3E. Print

a space after each seat, including after the last. Ex: numRows = 2 and numColumns = 3 prints:

1 answer:

Arada [10]3 years ago

3 0

Answer:

#include <stdio.h>

int main(void)

{

int Rows = 2;

int Col = 3;

int row = 0;

char colm = 'A';

char var = 'A';

for (row = 1; row<=Rows; ++row){

for (colm = 0; colm<Col; colm++){

printf("%d", row);

printf("%c ", var + colm % 3);

}

}

++var;

}

}

printf("\n");

return 0;

}

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A 20.0 µF capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to

Sergeu [11.5K]

Answer:

a) Q_initial = 16 * 10^-3 C

b) V_1 = V_2 = (16/3) * 10^2 V

c) E = 64/15 J

d) dE = 32/15 J of decrease

Explanation:

Given:

- Capacitor 1, C_1 = 20.0 uF

- Capacitor 2, C_2 = 10.0 uF

- Charged with P.d V = 800 V

Find:

a) the original charge of the system,

(b) the ﬁnal potential difference across each capacitor

(c) the ﬁnal energy of the system

(d) the decrease in energy when the capacitors are connected.

Solution:

a)

- The initial charge in the circuit is the one carried by the first charged capacitor.

Q_initial = C_1*V

Q_initial = 20*10^-6 * 800

Q_initial = 16 * 10^-3 C

b)

- After charging the other capacitor, we know that the total charge is conserved among two capacitor:

Q_initial = Q_1 + Q_2

- We also know that potential difference across two capacitor is also same.

V_1 = V_2 = Q_1 / C_1 = Q_2 / C_2

- Using the two equations and solve for charge Q_2:

Q_2 = Q_1*C_2/C_1

Q_2 = Q_1*10/20 = 0.5*Q_1

- using conservation of charge:

Q_initial = 1.5*Q_1

Q_1 = 16*10^-3 / 1.5 = 10.67*10^-3 C

- Hence the Voltage across each capacitor is:

V_2 = V_1 = Q_1 / C_1

= 10.67*10^-3 / 20*10^-6

= (16/3) * 10^2 V

c)

- The energy in the system is:

E = 0.5*C_eq*V^2

Where, C_eq is the equivalent capacitance of paralle circuit.

E = 0.5*(20+10)*10^-6 *((16/3) * 10^2)^2

E = 64/15 J

d)

- The decrease in energy of the capacitors is:

dE = E_initial - E_final

Where, E_initial is due to charging of the C_1 only:

dE = 0.5*10^-6*20*800^2 - (64/15)

dE = 32/5 - 64/15 = 32/15 J

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3 years ago

A barometer reads a height of 78 cmHg. Express this atmospheric pressure to:

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2 years ago

A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16

Bad White [126]

Answer:

the net work per cycle $\mathbf{W_{net} = 0.777593696}$ Btu per cycle

the power developed by the engine, W = 88.0144746 hp

Explanation:

the information given includes;

diameter of the four-cylinder bore = 3.7 in

length of the stroke = 3.4 in

The clearance volume = 16% = 0.16

The cylindrical volume $V_2 = 0.16 V_1$

the crankshaft N rotates at a speed of 2400 RPM.

At the beginning of the compression , temperature $T_1$ = 60 F = 519.67 R

and;

Otto cycle with a pressure = 14.5 lbf/in² = (14.5 × 144 ) lb/ft²

= 2088 lb/ft²

The maximum temperature in the cycle is 5200 R

From the given information; the change in volume is:

$V_1-V_2 = \dfrac{\pi}{4}D^2L$

$V_1-0.16V_1= \dfrac{\pi}{4}(3.7)^2(3.4)$

$V_1-0.16V_1= 36.55714291$

$0.84 V_1 =36.55714291$

$V_1 =\dfrac{36.55714291}{0.84 }$

$V_1 =43.52040823 \ in^3 \\ \\ V_1 = 43.52 \ in^3$

$V_1 = 0.02518 \ ft^3$

the mass in air ( lb) can be determined by using the formula:

$m = \dfrac{P_1V_1}{RT}$

where;

R = 53.3533 ft.lbf/lb.R°

$m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R \times 519 .67 ^0 R}$

m = 0.0018962 lb

From the tables of ideal gas properties at Temperature 519.67 R

$v_{r1} =158.58$

$u_1 = 88.62 Btu/lb$

At state of volume 2; the relative volume can be determined as:

$v_{r2} = v_{r1} \times \dfrac{V_2}{V_1}$

$v_{r2} = 158.58 \times 0.16$

$v_{r2} = 25.3728$

The specific energy $u_2$ at $v_{r2} = 25.3728$ is 184.7 Btu/lb

From the tables of ideal gas properties at maximum Temperature T = 5200 R

$v_{r3} = 0.1828$

$u_3 = 1098 \ Btu/lb$

To determine the relative volume at state 4; we have:

$v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}$

$v_{r4} =0.1828 \times \dfrac{1}{0.16}$

$v_{r4} =1.1425$

The specific energy $u_4$ at $v_{r4} =1.1425$ is 591.84 Btu/lb

Now; the net work per cycle can now be calculated as by using the following formula:

$W_{net} = Heat \ supplied - Heat \ rejected$

$W_{net} = m(u_3-u_2)-m(u_4 - u_1)$

$W_{net} = m(u_3-u_2- u_4 + u_1)$

$W_{net} = m(1098-184.7- 591.84 + 88.62)$

$W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)$

$W_{net} = 0.0018962 \times (410.08)$

$\mathbf{W_{net} = 0.777593696}$ Btu per cycle

the power developed by the engine, in horsepower. can be calculated as follows;

In the four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:

$W = 4 \times N' \times W_{net$

where ;

$N' = \dfrac{2400}{2}$

N' = 1200 cycles/min

N' = 1200 cycles/60 seconds

N' = 20 cycles/sec

W = 4 × 20 cycles/sec × 0.777593696

W = 62.20749568 Btu/s

W = 88.0144746 hp

8 0

2 years ago

A single fixed pulley is used to lift a load of 400N by the application of an effort of 480N in 10s through a vertical height of

Allushta [10]

Answer:

(a) the velocity ratio of the machine (V.R) = 1

(b) The mechanical advantage of the machine (M.A) = 0.833

(c) The efficiency of the machine (E) = 83.3 %

Explanation:

Given;

load lifted by the pulley, L = 400 N

effort applied in lifting the, E = 480 N

distance moved by the effort, d = 5 m

(a) the velocity ratio of the machine (V.R);

since the effort applied moved downwards through a distance of d, the load will also move upwards through an equal distance 'd'.

V.R = distance moved by effort / distance moved by the load

V.R = 5/5 = 1

(b) The mechanical advantage of the machine (M.A);

M.A = L/E

M.A = 400 / 480

M.A = 0.833

(c) The efficiency of the machine (E);

$E = \frac{M.A}{V.R} \times 100\%\\\\E = 0.833 \ \times \ 100\%\\\\ E = 83.3 \ \%$

4 0

2 years ago