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Zanzabum
3 years ago
6

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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I think because if you’ve already turned it in they might as well grade asap instead of waiting
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3 years ago
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A 20-mm-diameter steel bar is to be used as a torsion spring. If the torsional stress in the bar is not to exceed 110 MPa when o
ch4aika [34]

Answer:

1.887 m

Explanation:

(15 *pi)/180

= 0.2618 rad

Polar moment

= Pi*d⁴/32

= (22/7*20⁴)/32

= 15707.96

Torque on shaft

= ((22/7)*20³*110)/16

= 172857.14

= 172.8nm

Shear modulus

G = 79.3

L = Gjθ/T

= 79.3x10⁹x(1.571*10^-8)x0.2618/172.8

= 1.887 m

The length of the bar is therefore 1.887 meters

5 0
3 years ago
How do you know which forces works for free bodies​
miss Akunina [59]

Answer:

Gravitational force (pulled downward by the Earth)

Normal force (pushed upward by the ground)

Applied force (pushed by the person)

Friction force (pulled opposite the direction of motion by the roughness of the ground)

5 0
2 years ago
Four common causes of product failure are poor design, poor construction, poorly communicated operating instructions, and ______
Inga [223]

Answer:

B. operator error or misuse

Explanation:

A product is a failure if it is not able to achieve the anticipated life cycle as expected by the organization.

In such a case, there is a withdrawal of the product from the market as a result of its ultimate failure of a product to achieve profitability.

Four common causes of product failure are poor design, poor construction, poorly communicated operating instructions, and <u>operator error or misuse</u>

3 0
3 years ago
An axial compressive load of 708 kN is applied to a cylindrical component, 81 mm in diameter and 418 mm long, made of aluminium.
dalvyx [7]

Answer:

The compressive stress of aplying a force of 708 kN in a 81 mm diamter cylindrical component is 0.137 kN/mm^2 or 137465051 Pa (= 137.5 MPa)

Explanation:

The compressive stress in a cylindrical  component can be calculated aby dividing the compressive force F to the cross sectional area A:

fc= F/A

If the stress is wanted in Pascals (Pa), F and A must be in Newtons and square meters respectively.

For acylindrical component the cross sectional area A is:

A=πR^

If the diameter of the component is 81 mm, the radius is the half:

R=81mm /2 = 40.5 mm

Then A result:

A= 3.14 * (40.5 mm)^2  = 5150.4 mm^2

In square meters:

A= 3.14 * (0.0405 m)^2  = 0.005150 m^2

Replacing 708 kN to the force:

fc= 708 kN / 5150.4 mm^2 = 0.137 kN/mm^2

Using the force in Newtons:

F= 70800 N

Finally the compressive stress in Pa is:

fc= 708000 / 0.005150 m^2 = 137465051 Pa = 137 MPa

4 0
3 years ago
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