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Sveta_85 [38]
3 years ago
15

A positive rod is touched to a neutral sphere. There is a bar with positively charged particles attached to a ball with positive

and negative charged particles. Which best explains the arrow’s purpose in the diagram? It is showing the direction that protons move in, making the sphere negative. It is showing the direction that protons move in, making the sphere positive. It is showing the direction that electrons move in, making the sphere positive. It is showing the direction that electrons move in, making the sphere negative.

Physics
1 answer:
solong [7]3 years ago
4 0

Answer: A is correct

It is showing the direction that protons move in, making the sphere negative

Explanation:

From the figure, since the charges on the rod are completely positive, it shows that the protons are transferring from the sphere to the rod leaving the rod negatively charged.

Therefore, It is showing the direction that protons move in, making the sphere negative.

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A two-slit pattern is viewed on a screen 1.26 m from the slits. If the two fourth-order maxima are 53.6 cm apart, what is the to
Anettt [7]

Answer:

= 6.55cm

Explanation:

Given that,

distance = 1.26 m

distance between  two fourth-order maxima = 53.6 cm

distance between central bright fringe and fourth order maxima

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  = 26.8 cm

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tan θ = y / d

         = 0.268 m /  1.26 m

         = 0.2127

       θ = 12°

4th maxima

d sinθ = 4λ

d / λ = 4 / sinθ

d / λ = 4 / sin 12°

d / λ = 19.239

for first (minimum)

d sinθ = λ / 2

sinθ =  λ / 2d

       =  1 / 2(19.239)

       = 1 / 38.478

       = 0.02599

    θ =  1.489°

tan θ = y / d

y = d tan θ

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the total width of the central bright fringe  

Y = 2y

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4 0
3 years ago
When an aluminum bar is connected between a hot reservoir at 860 K and a cold reservoir at 348 K, 2.40 kJ of energy is transferr
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Answer:

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Explanation:

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\Delta S_{in} = \frac{2400\,J}{860\,K}

\Delta S_{in} = 2.791\,\frac{J}{K}

b) The change in entropy of the cold reservoir is:

\Delta S_{out} = \frac{2400\,J}{348\,K}

\Delta S_{out} = 6.897\,\frac{J}{K}

c) The total change in entropy of the Universe is modelled after the Second Law of Thermodynamics. Let assume that process is steady:

\Delta S_{in} - \Delta S_{out} + S_{gen} = 0

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6 0
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Answer:

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Total distance during the time interval  = 

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