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2 years ago

How would I find the Voltage across the open circuit

Engineering

1 answer:

Nimfa-mama [501]2 years ago

3 0

Answer:

Vab = 80V

Explanation:

The only current flowing in the circuit is supplied by the 100 V source. Its only load is the 40+60 ohm series circuit attached, so the current in that loop is (100V)/(40+60Ω) = 1A. That means V1 = (1A)(60Ω) = 60V.

Vab will be the sum of voltages around the right-side "loop" between terminals 'a' and 'b'. It is (working clockwise from terminal 'b') ...

Vab = -10V +60V +(0A×10Ω) +30V

Vab = 80V

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Which of the following refers to a full-scale version of a product used to validate performance?

kompoz [17]

I’m thinking it would be c sorry if it’s wrong .

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3 years ago

Read 2 more answers

The period of a pendulum T is assumed to depend only on the mass m, the length of the pendulum `, the acceleration due to gravit

zzz [600]

Answer:

The expression is shown in the explanation below:

Explanation:

Thinking process:

Let the time period of a simple pendulum be given by the expression:

$T = \pi \sqrt{\frac{l}{g} }$

Let the fundamental units be mass= M, time = t, length = L

Then the equation will be in the form

$T = M^{a}l^{b}g^{c}$

$T = KM^{a}l^{b}g^{c}$

where k is the constant of proportionality.

Now putting the dimensional formula:

$T = KM^{a}L^{b} [LT^{-} ^{2}]^{c}$

$M^{0}L^{0}T^{1} = KM^{a}L^{b+c}$

Equating the powers gives:

a = 0

b + c = 0

2c = 1, c = -1/2

b = 1/2

so;

a = 0 , b = 1/2 , c = -1/2

Therefore:

$T = KM^{0}l^{\frac{1}{2} } g^{\frac{1}{2} }$

T = $2\pi \sqrt{\frac{l}{g} }$

where k = $2\pi$

8 0

3 years ago

How can you evaluate whether the slope of the dependent variable with an independent variable is the same for each level of the

s2008m [1.1K]

Answer:

By running multiple regression with dummy variables

Explanation:

A dummy variable is a variable that takes on the value 1 or 0. Dummy variables are also called binary

variables. Multiple regression expresses a dependent, or response, variable as a linear

function of two or more independent variables. The slope is the change in the response variable. Therefore, we have to run a multiple regression analysis when the variables are measured in the same measurement.The number of dummy variables you will need to capture a categorical variable

will be one less than the number of categories. When there is no obvious order to the categories or when there are three or more categories and differences between them are not all assumed to be equal, such variables need to be coded as dummy variables for inclusion into a regression model.

3 0

3 years ago

A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po

Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

The distance of charge 2 from point B is

$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

$V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V$

So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

$W=q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

In this problem, we have:

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

4 0

3 years ago

A cantilever beam of length L = 70 in is made from two side-by-side structural-steel channels of size 3 in weighing 5.0 lbf/ft.

natali 33 [55]

Answer:

of 5 lb/ft and a concentrated service live load at midspan. .... length = 12 feet) to support a uniformly distributed load. Taking ... w 7..'{ 'f.- ~ s-·. 344 ft-kip. Fy : s-o ks I. 299 ft-kip. Li.. ::::- I 2.. }-t-. 150 ft-kip ..... The concrete and reinforcing steel properties are ... Neglecting beam self-weight . and based only on the ...... JI : Lf, 2. l.. ;VI.

Explanation:

6 0

3 years ago