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3 years ago

Batteries don’t run forever. Why do you think batteries run down?

Engineering

2 answers:

Marina86 [1]3 years ago

3 0

Because there is no such thing as immortal batteries.

When a battery is connected to a circuit, the charge moves through the circuit, and a chemical reaction occurs inside that separate the charges. The strength of this reaction diminishes over time and the battery eventually dies.

igomit [66]3 years ago

3 0

Answer;

Each charge cycle degrades the electrodes just a little bit more, meaning the battery loses performance over time, which is why even rechargeable batteries don't keep on working forever. Over the course of several charge and discharge cycles, the shape of the battery's crystals becomes less ordered.

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Write a C program that asks the user to enter two numbers, obtains the two numbers from the user and prints the sum, product, di

Bogdan [553]

Answer:

View Image

Explanation:

Initialize your variable as a float or double since you're going to be using fractions in your answer.

User scanf() to get user input.

Print out the sum, product, quotient, and difference between the two numbers.

8 0

3 years ago

A list is sorted in ascending order if it is empty or each item except the last one is less than or equal to its successor. HERE

Free_Kalibri [48]

Using the knowledge of computational language in python it is possible to write a code that writes a list and defines the arrange.

<h3>Writing code in python:</h3>

def isSorted(lyst):

if len(lyst) >= 0 and len(lyst) < 2:

return True

else:

for i in range(len(lyst)-1):

if lyst[i] > lyst[i+1]:

return False

return True

def main():

lyst = []

print(isSorted(lyst))

lyst = [1]

print(isSorted(lyst))

lyst = list(range(10))

print(isSorted(lyst))

lyst[9] = 3

print(isSorted(lyst))

main()

See more about python at brainly.com/question/18502436

#SPJ1

7 0

2 years ago

A helical compression spring is made with oil-tempered wire with wire diameter of 0.2 in, mean coil diameter of 2 in, a total of

Naya [18.7K]

Answer:

a. Solid length Ls = 2.6 in

b. Force necessary for deflection Fs = 67.2Ibf

Factor of safety FOS = 2.04

Explanation:

Given details

Oil-tempered wire,

d = 0.2 in,

D = 2 in,

n = 12 coils,

Lo = 5 in

(a) Find the solid length

Ls = d (n + 1)

= 0.2(12 + 1) = 2.6 in Ans

(b) Find the force necessary to deflect the spring to its solid length.

N = n - 2 = 12 - 2 = 10 coils

Take G = 11.2 Mpsi

K = (d^4*G)/(8D^3N)

K = (0.2^4*11.2)/(8*2^3*10) = 28Ibf/in

Fs = k*Ys = k (Lo - Ls )

= 28(5 - 2.6) = 67.2 lbf Ans.

c) Find the factor of safety guarding against yielding when the spring is compressed to its solid length.

For C = D/d = 2/0.2 = 10

Kb = (4C + 2)/(4C - 3)

= (4*10 + 2)/(4*10 - 3) = 1.135

Tau ts = Kb {(8FD)/(Πd^3)}

= 1.135 {(8*67.2*2)/(Π*2^3)}

= 48.56 * 10^6 psi

Let m = 0.187,

A = 147 kpsi.inm^3

Sut = A/d^3 = 147/0.2^3 = 198.6 kpsi

Ssy = 0.50 Sut

= 0.50(198.6) = 99.3 kpsi

FOS = Ssy/ts

= 99.3/48.56 = 2.04 Ans.

7 0

3 years ago

How many trips would one rubber-tired Herrywampus have to make to backfill a space with a geometrical volume of 5400 cubic yard?

nikklg [1K]

Answer:

If analyzed by volume capacity, more trips are needed to fill the space, thus the required trips are 288

Explanation:

a) By volume.

The shrinkage factor is:

$\frac{5400cu-yd}{1-0.25} =7200cu-yd$

The volume at loose is:

$V_{loose} =V_{bank} (1+swell-factor)=7200(1+0.2)=8640cu-yd$

If the Herrywampus has a capacity of 30 cubic yard:

$\frac{8640cu-yd}{30cu-yd/trip} =288trip$

b) By weight

The swell factor in terms of percent swell is equal to:

$pounds-per-cubic-yard-loose=\frac{pounds-per-cubic-yard-bank}{\frac{percent-swell}{100}+1 }$

$pounds-per-cubic-yard-loose=\frac{3000}{\frac{20}{100} +1} =2500lb/cu-yd$

The weight of backfill is:

$8640cu-yd*2500\frac{lb}{cu-yd} *\frac{1ton}{2000lb} =10800ton$

The Herrywampus has a capacity of 40 ton:

$\frac{10800}{40ton/trip} =270trip$

If analyzed by volume capacity, more trips are needed to fill the space, thus the required trips are 288

8 0

3 years ago

Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 440°C.

lawyer [7]

Answer:

Heat required (q) = 471.19kj/kg

Explanation:

Find attached below solution to problem

7 0

3 years ago