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disa
11 days ago
12

Which circuits correctly show Ohm's law?

Physics
2 answers:
Maru [2.9K]11 days ago
6 0
Ohm's Law is expressed as: V = IR where V denotes battery voltage, I represents current, and R indicates resistance. For the top-left circuit: V = 25V, I = 5A, R = 5ohm results in 25 = 5•5, which holds true. For the top-right circuit: V = 18V, I = 6A, R = 4ohm gives us 18 = 6•4, but 18 does not equal 24, making this circuit incorrect. The bottom-left circuit has V = 56V, I = 7A, R = 8ohm leading to 56 = 7•8, validating its correctness. Lastly, the bottom-right circuit: V = 20V, I = 5A, R = 4ohm where 20 = 5•4 also proves correct. Thus, the circuits that are accurate include the top-left, bottom-left, and bottom-right.
Sav [2.7K]11 days ago
5 0
The circuits that correctly demonstrate Ohm's law are the top left, bottom left, and bottom right.
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1 month ago
The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.939 W/m2. The wave is incident u
Yuliya22 [2947]

Answer:

The total energy can be expressed as T = 169.02 \ J

Explanation:

The problem states that

The Poynting vector, which measures energy flux, equals k = 0.939 \ W/m^2

The rectangle's length is represented by l = 1.5 \ m

The width of the rectangle is w = 2.0 \ m

The duration considered is t = 1 \ minute = 60 \ s

Mathematically, the overall electromagnetic energy incident on the area is given by

T = k * A * t

where A denotes the area of the rectangle, calculated as

A= l * w

By plugging in the respective values

A= 2 * 1.5

A= 3 \ m^2

Again substituting values

T = 0.939 * 3 * 60

T = 169.02 \ J

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1 month ago
Consider the uniform electric field E = (2.5 j + 3.5 k) × 103 N/C. (a) Calculate the electric flux through a circular area of ra
kicyunya [2889]
The desired electric flux Φ is calculated from the electric field E=(2.5• j + 3.5• k) ×10³ N/C and a circular path with a radius r=2.5m. The electric flux across a surface is represented as Φ=∮E•dA. Given the area A lies in the yz-plane, the normal orientation flows in the x-direction with A=πr² leading to dA=2πrdr •i. Thus, Φ evolves to Φ=∮(2.5j + 3.5k)×10³•(2πrdr i). Integrating from r=0 to r=2.5m and noting that components in different directions yield zero, results in Φ equaling 0 Nm²/C. Regarding the second part, when the area vector is at a 45° angle to the xy-plane, we redefine dA as (2πrCos45 i + 2πrSin45 j) dr, leading to the new flux calculation as Φ=10³∮ 5πrSin45 dr, integrating from 0 to 2.5m. With substitutions made, the result comes to Φ=34.71×10³ Nm²/C.
3 0
21 day ago
True or False: Molecules in a gas resist crowding and get as far apart as possible. Free electrons also resist crowding and get
ValentinkaMS [3034]

Answer:

This assertion is inaccurate.

Explanation:

The random nature of gas molecules results in their erratic motion and occasional collisions. While it is true that they tend to avoid being tightly packed, achieving the maximum separation from each other is not always feasible due to their lack of fixed positions. Consequently, gas molecules in a container cannot consistently maintain the furthest distance from their neighboring molecules.

In contrast, the separation among electrons is primarily influenced by repulsive forces, not random movement as in gases. Electrons maintain distance as a result of repulsion between similarly charged particles. Therefore, the arrangement of electrons on a charged copper sphere occurs not from a random distribution but rather due to repulsion, establishing a set distance between them.

4 0
25 days ago
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serg [3200]

Response:

0.9 cm

Clarification:

The following illustrates the calculation of the combined rod's length increase:

As established

Length increase = expansion of aluminum rod + expansion of steel rod

= 10cm \times 2.4e - 5\times (90-15) + 80cm\times 1.2e - 5\times (90-15)

= 0.9 cm

We simply summed the expansions of both the aluminum and steel rods to determine the overall increase in the joined rod's length, which must be factored in

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