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Katarina [22]
3 years ago
14

The cubic capacity of a four-stroke over-square spark-ignition engine is 245cc. The over-square ratio is (1.1) The clearance vol

ume is 27.2cc. Caleulate the bore, stroke, and compression ratio of the engine.
Engineering
1 answer:
pashok25 [27]3 years ago
4 0

Answer:

Bore = 7 cm

stroke = 6.36 cm

compression ratio = 10.007

Explanation:

Given data:

Cubic capacity of the engine, V = 245 cc

Clearance volume, v = 27.2 cc

over square-ratio = 1.1

thus,

D/L = 1.1

where,

D is the bore

L is the stroke

Now,

V = \frac{\pi}{4}D^2L

or

V = \frac{\pi}{4}\frac{D^3}{1.1}

on substituting the values, we have

245 =  \frac{\pi}{4}\frac{D^3}{1.1}

or

D = 7.00 cm

Now,

we have

D/L = 1.1

thus,

L = D/1.1

L = 7/1.1

or

L= 6.36 cm

Now,

the compression ratio is given as:

\textup{compression ratio}=\frac{V+v}{v}

on substituting the values, we get

\textup{compression ratio}=\frac{245+27.2}{27.2}

or

Compression ratio = 10.007

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What was the role of the rotors in the enigma machine?.
Lisa [10]

Answer:

Implements a reordering of the letters of the alphabet.

Explanation: GIVE ME 5 STARS AND a HEART!!!                                      Those contacts are wired across the rotor so that each contact on the left connects to the contact on the right in some scrambled arrangement. Each rotor, therefore, implements a reordering of the letters of the alphabet, which mathematicians call a permutation.

3 0
2 years ago
Koch traded Machine 1 for Machine 2 when the fair market value of both machines was $60,000. Koch originally purchased Machine 1
Mariana [72]

Answer:

Koch's adjusted basis in machine 2 after the exchange is $60,000

Explanation:

given data

fair market value = $60,000

originally purchased Machine 1 = $76,900

Machine 1 adjusted basis = $40,950

Machine 2 seller purchase = $64,050

Machine 2 adjusted basis = $55,950

solution

As he exchanged machine for another at $60,000

and this exchanged in fair market

so adjusted basis =  $50,000

Adjusted basis is the price of the item that affects the factors that are considered price. These factors usually include taxes, depreciation value, and other costs of acquiring and maintaining a given item. Adjusted basis is important so the right amount to sell

Adjusted basis increases when a person deducts expenses from factor taxes and operating statements

so Koch's adjusted basis in machine 2 after the exchange is $60,000

3 0
3 years ago
An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e
Mila [183]

Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in 2.25 * 10^4 BTU. It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.  

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - 1 BTU = 778.17  lbf*ft

2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:  

Ep = m*g*(h_2 - h_1)\\ W = m*g  

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.  

h_2 is the final elevation and h_1 is the initial elevation.

Replacing W in the Ep equation

Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\

Finally, the next step is to replace the variables of the problem.  

h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft

The elevation at the high point of the road is 12186.5 in ft.  

3 0
3 years ago
Write a method called letterCount that takes two String arguments, one containing some text and the other containing a single le
jenyasd209 [6]

Answer:

I am writing a Python program.  Here is the function letterCount. This function takes two string arguments text and letter and return count of all occurrences of a letter in the text.

def letterCount(text, letter):  

 count = 0  # to count occurrences of letter in the text string

 for char in text:  # loop moves through each character in the text

   if letter == char: # if given letter matches with the value in char

     count += 1  # keeps counting occurrence of a letter in text

 return count # returns how many times a letter occurred in text

   

Explanation:

In order to see if this function works you can check by calling this function and passing a text and a letter as following:

print(letterCount('apples are tasty','a'))

Output:

3

Now lets see how this function works using the above text and letter values.

text = apples are tasty

letter = a

So the function has to compute the occurrences of 'a' in the given text 'apples are tasty'.

The loop has a variable char that moves through each character given in the text (from a of apples to y of tasty) so it is used as an index variable.

char checks each character of the text string for the occurrence of letter a.

The if condition checks if the char is positioned at a character which matches the given letter i.e. a. If it is true e.g if char is at character a of apple so the if condition evaluates to true.

When the if condition evaluates to true this means one occurrence is found and this count variable counts this occurrence. So count increments every time the occurrence of letter a is found in apples are tasty text.

The loop breaks when every character in text is traversed and finally the count variable returns all of the occurrences of that letter (a) in the given text (apples are tasty). As a occurs 3 times in text so 3 is returned in output.

The screen shot of program along with output is attached.

4 0
2 years ago
Why is the definition for second based on the Cesium 133 atom?
jarptica [38.1K]

Answer:

  The rate of vibrations of the atoms is consistent

Explanation:

A second is defined as 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom.

The definition applies when the atom is in isolation at a temperature of absolute zero. Adjustments must be made for the conditions that can be maintained in practice.

Such radiation is determined by the laws of quantum mechanics, so is extremely consistent.

7 0
3 years ago
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