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3 years ago

Write a Nested While Loop that will increment the '*' from 1 to 10.

Engineering

1 answer:

andrey2020 [161]3 years ago

8 0

Answer:

The program is as follows:

i = 1

while(i<11):

j = 1

while(j<=i):

print('*', end = '')

j += 1

i += 1

print()

Explanation:

Initialize i to 1

i = 1

The outer loop is repeated as long as i is less than 11

while(i<11):

Initialize j to 1

j = 1

The inner loop is repeated as long as j is less than or equal i

while(j<=i):

This prints a *

print('*', end = '')

This increments j and ends the inner loop

j += 1

This increments i

i += 1

This prints a blank and ends the inner loop

print()

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Explain the working principles of a rotary engine

Brut [27]

Answer:

The rotary engine works on the same basic principle as the piston engine: combustion in the power plant releases energy to power the vehicle. However, the delivery system in the rotary engine is wholly unique. The piston engine performs four key operations: intake, compression, combustion, and exhaust.

Explanation:

7 0

3 years ago

Convert the following pairs of voltage and current waveforms to phasor form. Each pair of waveforms corresponds to an unknown el

exis [7]

Answer:

a) V = 20 ∠30⁰ , I = 4 ∠-210⁰ Z inductive L = 0,0125 H

b) V = 9∠-60⁰ , I = 4 ∠ 190⁰ Z capacitive C = 4,94 *10⁻⁴ F

c) V = 13 ∠240⁰ , I = 7 ∠ 150⁰ Z Inductive L = 0,0074 H

Explanation:

a) v(t) = 20 cos (400*t + 30 )

Phasor form V = 20 ∠30⁰

i(t) = 4 sin (400*t - 120)

First we need to transform 4sin( 400t - 120 ) as function cosine

we know that sin ( x + 90 ) = cos x

Then sin ( 400*t -120 ) = cos ( 400*t - 120 -90 ) = cos ( 400t - 120 - 90)

Phasor form I = 4 ∠-210⁰

To have the impedance nature we compute

Z = V / I ⇒ Z = 20 ∠30⁰ / 4 ∠-210⁰ Z = 5 ∠-180⁰

We notice that voltage advances the current then we are in presence of an inductive impedance

5 = wl ⇒ 5 = 400 *L ⇒ L = 0,0125 H

b) v(t) = 9 cos ( 900t - 60 )

V = 9∠-60⁰

i(t) = 4 sin ( 900t + 280 ) ⇒ i(t) = 4 cos ( 900t + 280 - 90)

i(t) = 4 cos (900t + 190 ) ⇒ I = 4 ∠ 190⁰

Z = V/I ⇒ Z = 9∠-60⁰ / 4 ∠ 190⁰ Z = 2,25 ∠-250

In this case the current advances the voltage. Impedance capacitive

1/wc = 1/ 900*C 1/wc = Z ⇒ 2,25 = 1/ 900*C

2,25*900 = 1/C ⇒ 2025 =1/C ⇒ C = 4,94 *10⁻⁴ F

c) v(t) = - 13 cos ( 250t + 60 )

v(t) = 13 cos ( 250t + 60 +180 ) ⇒ v(t) = 13 cos ( 250t +240)

Phasor Form

V = 13 ∠240⁰

i(t) = 7 sin (250t + 240 - 90) ⇒ i(t) = 7 sin (250t + 150)

Phasor Form I = 7 ∠150⁰

Z = 13∠240⁰ / 7 ∠150⁰ ⇒ Z = 1,86 ∠ 90⁰

Voltage advances the current then the impedance is inductive

wl = 250L 250 L = 1,86 L = 1,86/250 L = 0,0074 H

7 0

3 years ago

Elizabeth has been asked to design a pool umbrella that can withstand wind speeds of up to 30 miles per hour. Now that she under

zlopas [31]

Answer:

The correct option is;

D. She will need to research the problem even more

Explanation:

In other to properly design the pool umbrella that will be reliable and meet the needs of the particular customer, she will need to carry out a basic research about the environment where the pool umbrella is to be used, the presence of other factors that can impact on an idea for the pool umbrella.

She can then proceed to create a concept about the proposed pool umbrella, which is in the form of a paper concept

She will then proceed to carry out a more detailed study based on the paper concept that will consist of concept technology and application as well as functionality.

7 0

4 years ago

Read 2 more answers

The A-36 steel pipe has a 6061-T6 aluminum core. It issubjected to a tensile force of 200 kN. Determine the averagenormal stress

sasho [114]

Answer:

In the steel: 815 kPa

In the aluminum: 270 kPa

Explanation:

The steel pipe will have a section of:

A1 = π/4 * (D^2 - d^2)

A1 = π/4 * (0.8^2 - 0.7^2) = 0.1178 m^2

The aluminum core:

A2 = π/4 * d^2

A2 = π/4 * 0.7^2 = 0.3848 m^2

The parts will have a certain stiffness:

k = E * A/l

We don't know their length, so we can consider this as stiffness per unit of length

k = E * A

For the steel pipe:

E = 210 GPa (for steel)

k1 = 210*10^9 * 0.1178 = 2.47*10^10 N

For the aluminum:

E = 70 GPa

k2 = 70*10^9 * 0.3848 = 2.69*10^10 N

Hooke's law:

Δd = f / k

Since we are using stiffness per unit of length we use stretching per unit of length:

ε = f / k

When the force is distributed between both materials will stretch the same length:

f = f1 + f2

f1 / k1 = f2/ k2

Replacing:

f1 = f - f2

(f - f2) / k1 = f2 / k2

f/k1 - f2/k1 = f2/k2

f/k1 = f2 * (1/k2 + 1/k1)

f2 = (f/k1) / (1/k2 + 1/k1)

f2 = (200000/2.47*10^10) / (1/2.69*10^10 + 1/2.47*10^10) = 104000 N = 104 KN

f1 = 200 - 104 = 96 kN

Then we calculate the stresses:

σ1 = f1/A1 = 96000 / 0.1178 = 815000 Pa = 815 kPa

σ2 = f2/A2 = 104000 / 0.3848 = 270000 Pa = 270 kPa

5 0

3 years ago

Two vertical, parallel clean glass plates are spaced a distance of 2mm apart. if the plates are placed in water, how high will t

Ulleksa [173]

Answer with Explanation:

The capillary rise in 2 parallel plates immersed in a liquid is given by the formula

$h=\frac{2\sigma cos(\alpha )}{\rho gd}$

where

$\sigma$ is the surface tension of the liquid

$\alpha$ is the contact angle of the liquid

$\rho$ is density of liquid

'g' is acceleratioj due to gravity

'd' is seperation between thje plates

Part a) When the liquid is water:

For water and glass we have

$\sigma =7.28\times 10^{-2}N/m$

$\alpha =0$

$\rho _{w}=1000kg/m^3$

Applying the values we get

$h=\frac{2\times 7.28\times 10^{-2}cos(0)}{1000\times 9.81\times 2\times 10^{-3}}=7.39mm$

Part b) When the liquid is mercury:

For mercury and glass we have

$\sigma =485.5\times 10^{-3}N/m$

$\alpha =138^o$

$\rho _{w}=13.6\times 10^{3}kg/m^3$

Applying the values we get

$h=\frac{2\times 485.5\times 10^{-3}cos(138)}{13.6\times 1000\times 9.81\times 2\times 10^{-3}}=-2.704mm$

The negative sign indicates that there is depression in mercury in the tube.

4 0

3 years ago