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s344n2d4d5 [400]

4 years ago

11

2) Write a function called upsidedown_pyramid(height) that acceptsa parameter ""height"". It then prints an upside-down pyramid

of that height.(30

1 answer:

DiKsa [7]4 years ago

7 0

Hello, you haven't provided the programing language in which you need the code, I'll explain how to do it using Python, and you can follow the same logic to make a program in the programing language that you need.

Answer:

# -*- coding: utf-8 -*-

#Python

def upsidedown_pyramid(height):

maximun = (2*(height-1))+1

for i in range(height): print(' '*i+'*'*(maximun-i*2)+' '*i)

Explanation:

Define your function with an input parameter
Define your maximum, because you are not printing a normal pyramid but an upside-down pyramid you are going to start printing from the maximum number of '*' to just one, but what exactly is the maximum number of '*'? Because you build a pyramid starting with one block, '*', and each level you add two more blocks your formula to calculate the maximum is (2*(height-1))+1
Finally, you have to create a for loop to print each level. You need to start printing zero spaces and the maximum number of blocks, the next level you need two spaces and the maximum number of blocks minus two, and for each level that follows you are increasing by two the number of spaces, one left and one right, and decreasing by two the number of '*'

Note the constructing block of this pyramid is *

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Ti-6Al-4V has a fracture toughness of 74.6 MPa-m0.5. How much stress (in MPa) would it take to fail a plate loaded in tension th

Nikitich [7]

Answer:

critical stress = 595 MPa

Explanation:

given data

fracture toughness = 74.6 MPa- $\sqrt{m}$

crack length = 10 mm

f = 1

solution

we know crack length = 10 mm

and crack length = 2a as given in figure attach

so 2a = 10

a = 5 mm

and now we get here with the help of plane strain condition , critical stress is express as

critical stress = $\frac{k}{f\sqrt{\pi a}}$ ......................1

put here value and we get

critical stress = $\frac{74.6}{1\sqrt{\pi 5\times 10^{-3}}}$

critical stress = 595 MPa

so here stress is change by plane strain condition because when plate become thinner than condition change by plane strain to plain stress.

plain stress condition occur in thin body where stress through thickness not vary by the thinner section.

6 0

3 years ago

Which tool ensures that a fastener has the proper amount of tightness

Sidana [21]

A torque wrench tool is a tool that ensures that a fastener has the proper amount of tightness.

<h3>What is the torque wrench used for?</h3>

The torque wrench tool is used to ensure screws and bolts are properly tightened. When performing home repairs and maintenance of equipment it is quite important that a torque wrench is used in other to prevent a scenario where a fastener (screws and bolts) does not become loose leading to equipment failure or damage. Because of its many advantages, this tool is often found in the possession of construction workers.

You can learn more about the benefits of a torque wrench tool here

brainly.com/question/15075481

#SPJ1

7 0

2 years ago

How would you describe what would happen to methane if the primary bonds were to break?

erastova [34]

Answer:

All the bonds in methane (CH4CH4) are equivalent, and all have the same dissociation energy.

The product of the dissociation is methyl radical (CH3CH3). All the bonds in methyl radical are equivalent, and all have the same dissociation energy.

The product of that dissociation is methylene (CH2CH2). All the bonds in methylene are equivalent, and all have the same dissociation energy.

The product of that dissociation is methyne (CHCH) .

The C-H bonds in methane do not have the same dissociation energy as C-H bonds in methyl radical, which in turn do not have the same dissociation energy as the C-H bonds in methylene, which are again different from the C-H bond in methyne.

If (by some miracle) you were able to get all four bonds in methane to dissociate absolutely simultaneously, they would all show the same dissociation energy… but that energy, per bond broken, would be different than the energy required to break just one C-H bond in methane, because the products are different.

(In this case, it’s CH4→C+4HCH4→C+4H versus CH4→CH3+HCH4→CH3+H.)

To alter hydrocarbons you add enough energy to break a C-H bond. Why does only one bond break? What concentrates the energy on one C-H bond?

the weakest CH bond is the one that breaks. in plain alkanes it has to do with the molecular orbital interactions between neighboring carbon atoms. look at propane for example. the middle carbon has two C-C bonds, and each of those C-C bonds is strengthened by slight electron delocalization from the C-H bonds overlapping with the antibonding orbitals of the adjacent carbons.

since the C-H bonds on the middle carbon donate electron density to both of its neighbors, those two are weakest.

one of them will break preferentially.

which one actually breaks depends on the reaction conditions (kinetics). frankly it's whichever one ramdomly approaches a nucleophile first. when the nucleophile pulls of one of the H's, the other C-H bonds start to share (delocalize) the negative charge across the whole molecule. so while the middle C feels the majority of the negative charge character, the other two C's take on a fair amount as well...

by the way, alkanes don't really like to break and form anions like that.

a better example would be something like isopropyl iodide, where the C-I bond breaks and the I carries away the electron pair, forming a carbocation (also not particularly stable, but more so than the carbanion).

7 0

3 years ago

lisabon 2012 [21]

Answer:

That's a really nice question sadly I don't know the answer I'm replying to you cuz I'm tryna get points so... Sorry

3 0

4 years ago

The particle travels along the path defined by the parabola y=0.5x2, where x and y are in ft. If the component of velocity along

JulsSmile [24]

Answer:

D=41.48 ft

$a=54.43\ ft/s^2$

Explanation:

Given that

y=0.5 x²

Vx= 2 t

We know that

$V_x=\dfrac{dx}{dt}$

At t= 0 ,x=0

$x=\int V_x.dt$

At t= 3 s

$x=\int_{0}^{3} 2t.dt$

$x=[t^2\left\right ]_0^3$

x= 9 ft

When x= 9 ft then

y= 0.5 x 9² ft

y= 40.5 ft

So distance from origin is

x= 9 ft ,y= 40.5 ft

$D=\sqrt{9^2+40.5^2} \ ft$

D=41.48 ft

$a_x=\dfrac{dV_x}{dt}$

Vx= 2 t

$a_x= 2\ ft/s^2$

At t= 3 s , x= 9 ft

y=0.5 x²

$a_y=\dfrac{d^2y}{dt^2}$

y=0.5 x²

$\dfrac{dy}{dt}=x\dfrac{dx}{dt}$

$\dfrac{d^2y}{dt^2}=\left(\dfrac{dx}{dt}\right)^2+x\dfrac{d^2x}{dt^2}$

Given that

$\dfrac{dx}{dt}=2t$

$\dfrac{dx}{dt}=2\times 3$

$\dfrac{dx}{dt}=6\ ft/s$

$a_y=\dfrac{d^2y}{dt^2}=6^2+9\times 2\ ft/s^2$

$a_y=54\ ft/s^2$

$a=\sqrt{a_x^2+a_y^2}\ ft/s^2$

$a=\sqrt{2^2+54^2}\ ft/s^2$

$a=54.43\ ft/s^2$

7 0

4 years ago