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Makovka662
2 months ago
12

Two bars are conducting heat from a region of higher temperature to a region of lower temperature. The bars have identical lengt

hs and cross-sectional areas, but are made from different materials. In the drawing they are placed "in parallel" between the two temperature regions in arrangement A, whereas they are placed end to end in arrangement B. In which arrangement is the heat that is conducted the greatest?
Physics
1 answer:
inna [3.1K]2 months ago
3 0

Answer:

Arrangement A facilitates a greater rate of heat transfer

Explanation:

The rate of heat conduction between two materials differing in temperature adheres to the formula

\frac{\Delta Q}{\Delta t}=\frac{-KA \Delta T}{ L}

This equation demonstrates that

  • An enlarged equivalent cross-sectional area between the two materials will enhance the rate of heat transfer.
  • A reduction in the length of the medium will also amplify the rate of heat transfer

Arrangement A features a shorter transfer medium and a larger equivalent cross-sectional area, since the two rods are set up in a parallel arrangement.

Conversely, Arrangement B has a longer transfer medium and a smaller cross-sectional area, as the two rods are placed end to end

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Response:

To find power, we must first determine the work done by the force.

1) We will use the following equation to calculate work:

\int\limits {F} \, dx

The force is provided by the problem; our goal is to express 'dx' in terms of 't'

2) It's known that:

\frac{dV}{dt} = a = 2.6

Thus, we have:

v = 2.6t

Then:

\frac{dx}{dt} = V = 2.6t\\ \\dx = 2.6t*dt

3) Finally, substituting all known values gives us:

\int\limits^{4.7}_{0} {5.4t*2.6t} \, dt

After some calculations, the resulting work is:

161.9638 J.

4) To find power, we will use the following equation:

P = \frac{W}{t}

Thus

P = 161.9638/4.7 = 34.46 W

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2 months ago
Another element is silvery-white with a shiny luster, is very brittle, and forms ions with a −2 charge. give one possible identi
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The element is magnesium (Mg).
7 0
2 months ago
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How would the interference pattern change for this experiment if a. the grating was moved twice as far from the screen and b. th
Softa [3030]

Answer:

Refer to the explanation

Explanation:

Solution:-

- For this problem, we will assume the grating has a defined line density ( N ) representing the lines per mm.

- The angle formed by each fringe on the screen is given by ( θ ).

- The order of bright/dark spots is characterized by an integer ( n )

- The incident light's wavelength is ( λ )

- Using the diffraction grating relationship from the Young's Experiment is as follows:

                               n*lambda = \frac{sin(theta)}{N}

- The provided formula corresponds to constructive interference.

- We will examine how increasing the distance between the screen and the grating affects the pattern. Let ( L ) be the distance from the grating center to the screen center. The distance ( yn ) indicates the vertical spacing between each fringe on the screen.

- For small angles ( θ ), we use the approximation of sin ( θ ) ≈ tan ( θ ). Thus,

                            sin ( θ ) ≈ tan ( θ ) = [ yn / L ]

- Replacing this approximation in the diffraction grating relation results in:

                            y_n = n*lamda*L*N

- To double the distance from the screen to the grating, we use the relation with ( 2L ):

                            yn ∝ L

Result: The separation between each order of bright and dark fringe doubles. The interference pattern will spread further! Consequently, there will be fewer bright spots visible on the screen since a larger surface area is required to accommodate the entire pattern. The increased distance also diminishes the intensity contrast between bright and dark fringes due to the extended path traveled by light rays. Intensity is inversely related to the square of the travel distance.

- If the line density of the grating ( N ) were doubled, it follows that:

                            yn ∝ N

Result: The spacing between bright and dark fringes would also double. The interference pattern expands further, leading to the necessity of a bigger screen to show the entire pattern.

4 0
1 month ago
A hill is 132 m long and makes an angle of 12.0 degrees with the horizontal. As a 54 kg jogger runs up the hill, how much work d
inna [3103]

Answer:

14523.55J

Explanation:

The work exerted by the jogger against gravitational forces is represented by the following equation:

W=mgh.................(1)

where m signifies mass, g represents gravitational acceleration taken as 9.8m/s^2, and h denotes the elevation of the hill.

Given that the hill measures 132m in length and is inclined at an angle of 12 degrees to the horizontal, the height can be calculated as follows;

h=132sin12^o\\h=27.44m

Substitute this value into equation (1) along with all other required parameters to arrive at the result;

W=54*9.8*27.44\\W=14523.55J

4 0
2 months ago
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