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lana [24]
4 years ago
12

Calculate the resistance using Voltage and current, again using voltage and power, again using current and power, and again usin

g R1 and R2 recording the calculations for Run 3 rows 41-56
Engineering
1 answer:
ale4655 [162]4 years ago
6 0

Answer:

R = V / I ,   R = V² / P,     R = P / I²

Explanation:

For this exercise let's use ohm's law

      V = I R

      R = V / I

Electric power is defined by

      P = V I

ohm's law

      I = V / R

we substitute

      P = V (V / R)

      P = V² / R

      R = V² / P

 

the third way of calculation

      P = (i R) I

      P = R I²

      R = P / I²

You might be interested in
What’s the number of gold atoms in a nanogram? a picogram?
zvonat [6]

Answer :

The number of gold atoms in nanogram is, 3.057\times 10^{12}

The number of gold atoms in picogram is, 3.057\times 10^{9}

Explanation :

As we know that the molar mass of gold is, 196.97 g/mole. That means, 1 mole of gold has 196.97 grams of mass of gold.

As we know that,

1 mole contains 6.022\times 10^{23} number of atoms.

First we have to determine the number of gold atoms in a nanogram.

As, 196.97 grams of gold contains 6.022\times 10^{23} number of gold atoms

And, 1 grams of gold contains \frac{1g}{196.97g}\times (6.022\times 10^{23}) number of gold atoms

So, 10^{-9} nanograms of gold contains \frac{1g}{196.97g}\times (10^{-9})\times (6.022\times 10^{23})=3.057\times 10^{12} number of gold atoms

The number of gold atoms in nanogram is, 3.057\times 10^{12}

Now we have to determine the number of gold atoms in a picogram.

As, 196.97 grams of gold contains 6.022\times 10^{23} number of gold atoms

And, 1 grams of gold contains \frac{1g}{196.97g}\times (6.022\times 10^{23}) number of gold atoms

So, 10^{-12} picograms of gold contains \frac{1g}{196.97g}\times (10^{-12})\times (6.022\times 10^{23})=3.057\times 10^{9} number of gold atoms

The number of gold atoms in picogram is, 3.057\times 10^{9}

8 0
3 years ago
A smelter emits SO2 at a rate of 10,000 kg/day. The stack is 200 m tall and the plume rise is 100 m. The windspeed is 2 m/s at a
Olegator [25]

Answer:

lol

Explanation:

lol

5 0
3 years ago
The following statements are about the laminar boundary layer over a flat plate. For each statement, answer whether the statemen
Nikolay [14]

Answer:

1. B. False

2.  B. False

3. A. True

4. B. False

5. A. True

6. A. True

7. A. True

Explanation:

1. B. False

The relation of Reynolds' number, Reₓ to boundary layer thickness δ at a point x is given by the relation

\delta = \dfrac{x \times C}{\sqrt{Re_x} }

That is the boundary layer thickness is inversely proportional to the square root of the Reynolds' number so that if the Reynolds' number were to increase, the boundary layer thickness would decrease

Therefore, the correct option is B. False

2.  B. False

From the relation

Re_x = \dfrac{U_o \times x}{v}

As the outer flow velocity increases, the boundary layer thickness diminishes

3. A. True

As the viscous force is increased the boundary layer thickness increases

4. B. False

Boundary layer thickness is inversely proportional to velocity

5. A. True

The boundary layer model developed by Ludwig Prandtl is a special case of the Navier-Stokes equation

6. A. True

Given a definite boundary layer thickness, the curve representing the boundary layer thickness is a streamline

7. A. True

The boundary layer approximation by Prandtl Euler bridges the gap between the Euler (slip boundary conditions) and Navier-Stokes (no slip boundary conditions) equations.

6 0
3 years ago
Disk A has a mass of 8 kg and an initial angular velocity of 360 rpm clockwise; disk B has a mass of 3.5 kg and is initially at
garri49 [273]

Answer:

A. αa= 9.375 rad/s^2, αb= 28.57 rad/s^2

B. ωa=251rpm, ωb=333rpm

Explanation:

Mass of disk A, m1= 8kg

Mass of disk B, m2= 3.5kg

The initial angular velocity of disk A= 360rpm

The horizontal force applied = 20N

The coefficient of friction μk = 0.15

While slipping occurs, a frictional force is applied to disk A and disk B

T= 1/2Mar^2a

T=1/2 (8kg) (0.08)^2

T= 0.0256kg-m^2

N=P=20N

F= μN

F= 0.15 × 20

F=3N

We have

Summation Ma= Summation(Ma) eff

Fra= Iaαa

(3N)(0.08m)= (0.0256kg-m^2)α

αa= 9.375 rad/s^2

The angular acceleration at disk A is αa= 9.375 rad/s^2 is acting in the anti-clockwise direction.

For Disk B,

T= 1/2Mar^2a

T= 1/2(3.5) (0.06)^2

=0.0063kg-m^2

We have,

Summation Mb= Summation(Mb) eff

Frb= Ibαb

(3N)(0.06m)= (0.0063kg-m^2) α

αb= 28.57 rad/s^2

B) ( ωa)o= 360rpm(2 pi/60)

= 1 pi rad/s

The disk will stop sliding where

ωara=ωbrb

(ωao-at)ra=αtr

(12pi-9.375t) (0.08)=28.57t(0.06)

(37.704-9.375t)(0.08)= 1.7142t

3.01632-0.75t= 1.7142t

t=1.22s

Now,

ωa=(ωa)o- t

12pi - 9.375(1.22)

37.704-11.4375

=26.267 rad/s

26.267× (60/2pi)

= 250.80

251rpm

The angular velocity at a, ω= 251rpm

Now,

ωb= αbt

=28.57(1.22)

=34.856rad/s

34.856rad/s × (60/2pi)

=332.807

= 333rpm

Therefore the angular velocity at b ω=333rpm

4 0
3 years ago
Read 2 more answers
he mean weight of a breed of yearling cattle is 11871187 pounds. Suppose that weights of all such animals can be described by th
Tamiku [17]

Question

The mean weight of a breed of yearling cattle is 1187 pounds. Suppose that weights of all such animals can be described by the Normal model ​N(1187,78). ​

a) How many standard deviations from the mean would a steer weighing 1000 pounds​ be?

b) Which would be more​ unusual, a steer weighing 1000 ​pounds, or one weighing 1250 ​pounds? ​

Answer:

a. z = -2.40

A sleet weighing 1,000 pounds is 2.40 standard deviations below the mean.

b. z = 0.81

1000 is more unusual because its contained on the extreme end from the mean

Explanation:

a.

Let weight (in pounds) of the cattle be denoted by letter x:

z = (x - u)/ σ

Where u = mean and σ = standard deviation

u = 1187

σ = 78

x = 1000

Use z score formula to standardize the value of x:

z = (1000 - 1187)/78

z = -187/78

z = -2.397436

z = -2.40 ------_ Approximated

A sleet weighing 1,000 pounds is 2.40 standard deviations below the mean.

b.

x= 1250

z= (1250 - 1187)/78

z = 63/78

z = 0.807692

z = 0.81 --------- Approximated

1000 is more unusual because its contained on the extreme end from the mean

8 0
3 years ago
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