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4 years ago

A low-power microprocessor consumes 5 mA at DC. How many electrons flow through the power wiring per second?

Physics

1 answer:

Marta_Voda [28]4 years ago

4 0

Answer:

Explanation:

time, t = 1 second

i = 5 mA = 0.005 A

Charge of one electron, e = 1.6 x 10^-19 C

Charge, q = i t

q = 0.005 x 1 = 0.005 C

Number of electrons in one second = total charge / charge of one electron

$\frac{0.005}{1.6 \times 10^{-19}}=3.125 \times 10^{16}$

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What do organisms need in order to reproduce? This is Science Btw

Delicious77 [7]

Answer:

Energy. They need energy.

Explanation:

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3 years ago

Read 2 more answers

If you had 8 balls and 7 of them were a certain weight, and 1 of them was heavier, how could you find the heaviest ball. All the

dybincka [34]

Answer:

There are two method of comparing the balls 1) using a balance 2) by only 2 weighings.

Explanation:

There are two method of comparing the balls 1) using a balance 2) by only 2 weighings.

Make the following groups - --- (1,2,3),(4,5,6),(7,8)

Step 1. compare the Weigh (1,2,3) and (4,5,6)

there are 2 possible outcomes:

1---both the group are of same weight. and named as (Case A)

2--- one of the group is heavier than other and named as (Case B)

Step 2. Let examine both case

In Case A --in this case, now compare the weight of 7th and 8th ball. By this you have recognize the heavier ball by 2 weighing method.

In Case B -- considered the heaviest group (assume group (1,2,3) is heavy), from this group take randomly two ball and compare the their weight. out of these two ball, one is heavy else the third ball is.

7 0

3 years ago

One of the most efficient engines built so far has the following characteristics: combustion chamber temperature = 1900°C, exhau

suter [353]

Answer:

actual efficiency is 47.78 %

Carnot efficiency is 67.65 %

power output is 5.20 × 10^3 hp

Explanation:

given data

temperature = 1900°C = 1900+ 273 K = 2173 K

exhaust temperature = 430°C = 430 + 273 K = 703 K

fuel = 7.0 × 10^9 cal

work = 1.4 × 10^10 J

to find out

actual efficiency and Carnot efficiency and power output of engine

solution

first we find actual efficiency that is = work / heat input

put the value and

input energy = 7.0 × 10^9 cal (4.184 J/1 cal) = 29.29 × 10^9 J

actual efficiency = 1.4 × 10^10 / ( 29.29 × 10^9 )

actual efficiency = 0.4778

actual efficiency is 47.78 %

and

Carnot efficiency is = 1 - ( 703 / 2173 )

so Carnot efficiency is = 0.67648

Carnot efficiency is 67.65 %

and

power output = work / time

power output = 1.4 × 10^10 / 3600 sec

power output = 3.88 × 10^6 W

power output = 3.88 × 10^6 W / 746 hp

so power output is 5.20 × 10^3 hp

5 0

3 years ago

A box slides down a 30.0° ramp with an acceleration of 1.20 m/s^2. Determine the coefficient of kinetic friction between the box

Zielflug [23.3K]

m = mass of the box

N = normal force on the box

f = kinetic frictional force on the box

a = acceleration of the box

μ = coefficient of kinetic friction

perpendicular to incline , force equation is given as

N = mg Cos30 eq-1

kinetic frictional force is given as

f = μ N

using eq-1

f = μ mg Cos30

parallel to incline , force equation is given as

mg Sin30 - f = ma

mg Sin30 - μ mg Cos30 = ma

"m" cancel out

a = g Sin30 - μ g Cos30

inserting the values

1.20 = (9.8) Sin30 - (9.8) Cos30 μ

μ = 0.44

4 0

4 years ago

If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.69 rev/s , friction in

Sindrei [870]

Answer:

magnitude of the frictional torque is 0.11 Nm

Explanation:

Moment of inertia I = 0.33 kg⋅m2

Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s

Final angular velocity w = 0 (since it stops)

Time t = 13 secs

Using w = w° + §t

Where § is angular acceleration

O = 4.34 + 13§

§ = -4.34/13 = -0.33 rad/s2

The negative sign implies it's a negative acceleration.

Frictional torque that brought it to rest must be equal to the original torque.

Torqu = I x §

T = 0.33 x 0.33 = 0.11 Nm

5 0

4 years ago