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3 years ago

9

Two copper wires with different diameters are joined end to

2 answers:

Harman [31]3 years ago

7 0

The current remains same in an end to end (or series) connection ! so, with decrease in cross- sectional area, drift velocity must increase!They experience an "impedance bump" which will affect the total current flow in the circuit.So,current remains constant while th drift velocity increases in areas of less cross-section area..

xxTIMURxx [149]3 years ago

6 0

Decrease because the gauge of the wires are small

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A ball is projected upward at time t=0.0s, from a point on a roof 90m above the ground. The ball rises, then falls and strikes t

chubhunter [2.5K]

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3 years ago

A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate

diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

$W = Q_{source}-Q_{sink}$

According to the data given we have to,

$Q_{source} = 200000Btu$

$T_{source} = 1500R$

$Q_{sink} = 100000Btu$

$T_{sink} = 600R$

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

$\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}$

$\Delta S_{sink} = \frac{100000}{600}$

$\Delta S_{sink} = 166.67Btu/R$

On the other hand,

$\Delta S_{source} = \frac{Q_{source}}{T_{source}}$

$\Delta S_{source} = \frac{-200000}{1500}$

$\Delta S_{source} = -133.33Btu/R$

The total change of entropy would be,

$S = \Delta S_{source}+\Delta S_{sink}$

$S = -133.33+166.67$

$S = 33.34Btu/R$

Since $S\neq 0$ the heat engine is not reversible.

PART B)

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Therefore the work in the system is 100000Btu

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3 years ago

A 35 g steel ball is held by a ceiling-mounted electromagnet 3.8 m above the floor. A compressed-air cannon sits on the floor, 4

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Answer:

Explanation:

Let v is the launch speed of the plastic ball and the angle of projection is θ.

So, in horizontal direction

v Cosθ x t = 4.8 .... (1)

In th evertical direction

1.4 = v Sin θ x t - 0.5 gt² .... (2)

As , v Sin θ x t = 3.8 .... (3) , put in equation (2)

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v Sinθ = 5.43

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