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kati45 [8]
3 years ago
10

How long does it take electrons to get from a car battery to the starting motor? Assume the current is 300 A and the electrons t

ravel through a copper wire with cross-sectional area 0.21 cm2 and length 0.85 m.The number of charge carriers per unit volume is 8.49 1028 m3.
Engineering
1 answer:
yuradex [85]3 years ago
6 0

Answer:

t=13.49 min

Explanation:

Given that

I=300 A

A= 0.21 cm²

L= 0.85 m

Number of charge carriers per unit volume = 8.49 x 10²⁸ m⁻³.

We know speed of the electron given as

V=J/ne

J= I/A

So

V=I/neA

V=\dfrac{300}{8.49\times 10^{28}\times 1.6\times 10^{-19}\times 0.21\times 10^{-4} }

V=0.00105 m/s

lets take t is the time when electron travels from car battery to the starting motor

t = L/V

t=\dfrac{0.85}{0.00105}

t=809.52 s

t=13.49 min

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An engine has a piston with a surface area of 17.31 in2 and can travel 3.44 inches. What is the potential change in volume, disp
Katena32 [7]

Answer:

$$\begin{align*}

P(Y-X=m | Y > X) &= \sum_{k} P(Y-X=m, X=k | Y > X) \\ &= \sum_{k} P(Y-X=m | X=k, Y > X) P(X=k | Y > X) \\ &= \sum_{k} P(Y-k=m | Y > k) P(X=k | Y > X).\end{split}$$

Explanation:

\eqalign{

 P(Y-X=m\mid Y\gt X)

   &=\sum_kP(Y-X=m,X=k\mid Y\gt X)\cr

   &=\sum_kP(Y-X=m\mid X=k,Y\gt X)\,P(X=k\mid Y>X)\cr

   &=\sum_kP(Y-k=m\mid Y\gt k)\,P(X=k\mid Y\gt X)\cr

}

P(Y-X=m | Y > X) &= \sum_{k} P(Y-X=m, X=k | Y > X) \\ &= \sum_{k} P(Y-X=m | X=k, Y > X) P(X=k | Y > X) \\ &= \sum_{k} P(Y-k=m | Y > k) P(X=k | Y > X).\end{split}$$

5 0
2 years ago
Write a new ARMv8 assembly file called "lab04b.S" which is called by your main function. It should have the following specificat
Len [333]

Answer:

my_mul:

.globl my_mul

my_mul:

   //Multiply X0 and X1

   //   Does not handle negative X1!

   //   Note : This is an in efficient way to multipy!

   SUB SP, SP, 16       //make room for X19 on the stack

   STUR X19, [SP, 0]    //push X19

   ADD X19, X1, XZR     //set X19 equal to X1

   ADD X9 , XZR , XZR //set X9 to 0

mult_loop:

   CBZ X19, mult_eol

   ADD X9, X9, X0

   SUB X19, X19, 1

   B mult_loop

mult_eol:

   LDUR X19, [SP, 0]

   ADD X0, X9, XZR      // Move X9 to X0 to return

   ADD SP, SP, 16       // reset the stack

   BR X30

Explanation:

6 0
3 years ago
You are a designer of a new processor. You have to choose between two possible implementations (called M1 and M2) of the same ar
Kaylis [27]

Answer:

A ) CPI : M1 = 2.4 , M2 = 2.65

B ) MIPS : M1 = 1083, M2 = 1056

C ) The machine that has a better performance based on MIPS is M1 and this is  by 27 million number of instructions per sec

Explanation:

A) The CPI for each machine

CPI = ( Total number of execution cycles ) / ( instruction counter executed )

For Machine 1 ( M1 )

we have to make some assumptions : number of instructions = 10

number of times A was executed = 4 , Number of times B was executed = 2.5 , number of times C was executed = 2.5, Number of times D was executed = 1. and this was based on the frequency given above

hence CPI for M1 =[ ( 1 * 4 ) + ( 3 * 2.5 ) + ( 3 * 2.5 ) + ( 5 * 1 ) ] / 10

       CPI  for M1 = 2.4

For Machine 2 ( M2 )

we have to make some assumptions : number of instructions = 10

number of times A was executed = 4 , Number of times B was executed = 2.  number of times C was executed = 1.5, Number of times D was executed = 2.5 times. and this was based on the frequency given above

Hence CPI  for M1 = [ ( 2 * 4 ) + ( 2 * 2 ) + ( 3 * 1.5 ) + ( 4 * 2.5 ) ] / 10

            CPI for M2 = 2.65

B ) Calculate the native MIPS  ratings for M1 and M2

MIPS = ( instruction counts ) / ( Execution time * 10^6 )

For M1

Assumptions : number of instructions executed = 10

                        each clock cycle = 0.3846 * 10^-9.      frequency = 2.6 Ghz

first we calculate the total execution time which is equal to :

= [ ( 1 * 4 ) + ( 3 * 2.5 ) + ( 3 * 2.5 ) + ( 5 * 1 ) ] * 0.3846 * 10 ^-9

= 9.2304 * 10 ^-9 secs

therefore the MIPS for M1

= 10 / ( 9.2304 * 10^-9 ) * 10^6  = 1083

                                         

For M2

Assumptions : number of instructions executed = 10

                        each clock cycle = 0.3846 * 10^-9.      frequency = 2.8 Ghz

first we calculate the total execution time which is equal to :

= [ (2*4) + (2*2) + (3 * 1.5 ) + ( 4 * 2.5 ) ] * 0.3846 * 10^-9 = 9.4631 * 10^-9 secs

therefor the MIPS for M2

= 10 / ( 9.4631*10^-9) * 10^6 = 1056

C ) The machine that has a better performance based on MIPS is M1 and this is by 27 million number of instructions per sec

8 0
3 years ago
A circular ceramic plate that can be modeled as a blackbody is being heated by an electrical heater. The plate is 30 cm in diame
MakcuM [25]

Answer:

Heater power = 425 watts

Explanation:

Detailed explanation and calculation is shown in the image below

6 0
3 years ago
(Practice work, not graded)
luda_lava [24]

Answer: ALL CAREFULLY ANSWERED CORRECTLY.

Explanation:

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3) The theory of systems view the world as a complex system of interconnected parts. If we consider the society; (financial systems, political systems, etc) we will agree that they individually have their own components and it's the summation of this components that makes the system, this implies that system thinking could be applicable in this kinda of systems as long as they are made up of components.

4) Technology has boosted every sector of our lives and it has the capacity to do more. Restricting it's importance to entertainment alone would be an underusing of its potentials. Engineering students infact should not need any drive to be encouraged about maximizing all it can do in shaping our world.

5) ~ Nature shows its splendid soul

~Never ceases to leave us in amazement

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7 0
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