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URGENT!!!!!! Assume that a wire has 1.5 ohms of resistance. If the wire is connected to two batteries with a total voltage of 3.0 V, how much current will flow through the wire? 3.0 amps 2.3 amps 2.0 amps 1.0 amps

Answers

Answer 1

Answer:

Answer:

2.0 amps

Explanation:

Current is the ratio of voltage to resistance:

I = V/R = (3.0)/(1.5) = 2.0

The current in the wire is 2.0 amps.

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Part of your electrical load is a 60-W light that is on continuously. By what percentage can your energy consumption be reduced by turning this light off

Answers

Answer:

Following are the solution to the given question:

Explanation:

Please find the complete question in the attached file.

The cost after 30 days is 60 dollars. As energy remains constant, the cost per hour over 30 days will be decreased.

$\to (\$60)/((30 \ days)/(24\ hours)) = \$0.08 / kwh.$

Thus, $(\$0.08)/(\$0.12) = 0.694 \ kW * 0.694 \ kW * 1000 = 694 \ W.$

The electricity used is continuously 694W over 30 days.

If just resistor loads (no reagents) were assumed,

$\to I = (P)/(V)= (694\ W)/(120\ V) = 5.78\ A$

Energy usage reduction percentage = $((60\ W)/(694\ W) * 100\%)$

This bulb accounts for $8.64\%$ of the energy used, hence it saves when you switch it off.

True Or False for each question

Answers

7 true

8 false

9 false

10 false

11 false

12 true

13 true

hope this helps!

Can you guys please help me out

Answers

The answer is A I’m pretty sure it is

Which law does the following statement express? "In all cases of electromagnetic induction, the induced voltages have adirection such that the currents they produce oppose the effect that produces them."

Answers

Final answer:

Faraday's Law of electromagnetic induction states that induced voltages produce currents that oppose the change in the magnetic field.

Explanation:

The law that the statement expresses is Faraday's Law of electromagnetic induction.

According to Faraday's Law, whenever there is a change in the magnetic field through a conductor, it induces an electromotive force (EMF) or voltage across the conductor. This induced voltage creates a current that flows in a direction that opposes the change in magnetic field.

This phenomenon is described by Lenz's Law, which states that the induced current always flows in such a way as to produce a magnetic field that opposes the change in the external magnetic field.

Learn more about Faraday's Law of electromagnetic induction here:

brainly.com/question/13369951

The position of a particle as a function of time is given by x = (2.0 m/s)t + (-3.0 m/s2)t2. (a) plot x-versus-t for time from t = 0 to t = 1.0 s. (do this on paper. your instructor may ask you to turn in this plot.) this answer has not been graded yet. (b) find the average velocity of the particle from t = 0.45 s to t = 0.55 s. m/s (c) find the average velocity from t = 0.49 s to t = 0.51 s.

Answers

Part a)

Equation of position with time is given as

$x = (2.0 m/s)t + (-3.0 m/s2)t^2$

since this equation is a quadratic equation

so it will be a parabolic graph between t = 0 to t = 1

part b)

at t = 0.45 s

$x = 2* 0.45 - 3 * 0.45^2$

$x_1 = 0.2925 m$

at t = 0.55 s

$x = 2* 0.55 - 3*0.55^2$

$x_2 = 0.1925$

now the displacement is given as

$d = x_2 - x_1$

$d = 0.1925 - 0.2925 = -0.1 m$

so the average velocity is given by

$v = (d)/(t)$

$v = \frac{-0.1}{0.1) = -1 m/s$

part c)

at t = 0.49 s

$x = 2* 0.49 - 3 * 0.49^2$

$x_1 = 0.2597 m$

at t = 0.51 s

$x = 2* 0.51 - 3*0.51^2$

$x_2 = 0.2397 m$

now the displacement is given as

$d = x_2 - x_1$

$d = 0.2397 - 0.2597 = -0.02 m$

so the average velocity is given by

$v = (d)/(t)$

$v = \frac{-0.02}{0.02) = -1 m/s$

What electric field strength would store 12.5 JJ of energy in every 6.00 mm3mm3 of space?

Answers

To develop this problem we will apply the concepts related to the potential energy per unit volume for which we will obtain an energy density relationship that can be related to the electric field. From this formula it will be possible to find the electric field required in the problem. Our values are given as

The potential energy, $U = 13.0 J$

The volume, $V = 6.00 mm^3 = 6.00*10^(-9)m^3$

The potential energy per unit volume is defined as the energy density.

$u = (U)/(V)$

$u= ((13.0 J))/((6.00*10^(-9) m^3))$

$u= 2.167109 J/m^3$

The energy density related with electric field is given by

$u = (1)/(2) \epsilon_0 E^2$

Here, the permitivity of the free space is

$\epsilon_0 = 8.85*10^-{12} C^2/N \cdot m^2$

Therefore, rerranging to find the electric field strength we have,

$E = \sqrt{(2u)/(\epsilon_0)}$

$E = \sqrt{(2(2.167109))/(8.85*10^(-12))}$

$E = 2.211010 V/m$

Therefore the electric field is 2.21V/m

Final answer:

To calculate the electric field strength that would store 12.5 Joules of energy in every 6.00 mm^3 of space, we use the energy density formula. We firstly find the energy density and input it into the formula to solve for the electric field strength. The result is approximately 6.87 X 10^6 N/C.

Explanation:

The energy stored in an electric field is given by the formula U = 1/2 ε E^2. Here, U is the energy density (energy per unit volume), E is the electric field strength, and ε is the permittivity of free space.

Given that the energy stored U is 12.5 joules, and the volume is 6.00 mm^3 or 6.00X10^-9 m^3, the energy density (U) can be computed as 12.5 J/6.00X10^-9 m^3 = 2.08X10^12 Joule/meter^3.

We can solve the formula for E (electric field strength): E = sqrt ((2U)/ε). Substituting the value of ε (8.85 × 10^-12 m^-3 kg^-1 s^4 A^2), we can find E to be approximately 6.87 X 10^6 N/C.

Learn more about Electric Field Strength here:

brainly.com/question/1812671

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