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3 years ago

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If a current-carrying wire has a cross-sectional area that gradually becomes smaller along the length of the wire, the drift v

elocity:________
a. increases along the length of the wire if the resistance increases too
b. decreases along the length of the wire if the resistance decreases too
c. decreases along the length of the wire
d. increases along the length of the wire
e. remains the same along the length of the wire

1 answer:

jarptica [38.1K]3 years ago

6 0

Answer:d

Explanation:

Drift velocity is given by

$v_d=\frac{I}{n\times Q\times A}$

where $v_d$ =drift velocity

I=Current

n=no of electron

Q=charge of Electron

A=cross-section

If area of cross-section decreases gradually then drift velocity will increase because drift velocity is inversely proportional to Area of cross-section

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g = 9.8 m/s²

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∑F = F1 + W

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∑F = 20 + 29.4

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3 years ago

Read 2 more answers

A roller skater of 47kg moving with a velocity of 12 m/s to the east picks up a bag of 6.0 kg. What is the final velocity of the

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Answer:

v_f = 10.85 m/s

Explanation:

We will apply the law of conservation of momentum here:

$m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f}+m_{2}v_{2f}\\$

where,

m₁ = mass of roller skater = 47 kg

m₂ = mass of bag = 6 kg

v_1i = initial speed of roller skater = 12 m/s

v_2i = initial speed of the bag = 0 m/s

v_1f = final speed of the roller skater = ?

v_2f = final speed of the bag = ?

Both the bag and the skater will have same speed at the end because kater is carrying the bag:

v_1f = v_2f = v_f

Therefore, the equation will become:

$(47\ kg)(12\ m/s)+(6\ kg)(0\ m/s)=(47\ kg)(v_{f})+(5\ kg)(v_{f})\\564\ N.s = (47\ kg+5\ kg)(v_{f})\\v_{f} = \frac{564\ N.s}{52\ kg}\\$

<u>v_f = 10.85 m/s</u>

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2 years ago

A force of 44 N will stretch a rubber band 88 cm (0.080.08 m). Assuming that Hooke's law applies, how far will aa 11-N forc

Setler79 [48]

Answer:

<em>The rubber band will be stretched 0.02 m.</em>

<em>The work done in stretching is 0.11 J.</em>

Explanation:

Force 1 = 44 N

extension of rubber band = 0.080 m

Force 2 = 11 N

extension = ?

According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.

F = ke

where k = constant of elasticity

e = extension of the material

F = force applied.

For the first case,

44 = 0.080K

K = 44/0.080 = 550 N/m

For the second situation involving the same rubber band

Force = 11 N

e = 550 N/m

11 = 550e

extension e = 11/550 = <em>0.02 m</em>

<em>The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch</em>. This is in line with energy conservation.

potential energy stored = $\frac{1}{2}ke^{2}$

==> $\frac{1}{2}* 550* 0.02^{2}$ = <em>0.11 J</em>

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2 years ago

A capacitor with an initial potential difference of 185 V is discharged through a resistor when a switch between them is closed

GrogVix [38]

Answer:

a. $\tau = 2.1161 s$

b. $V(18.8 \ s) = 0.0256 \ V$

Explanation:

<h3>a.</h3>

The equation for the voltage V of discharging capacitor in an RC circuit at time t is:

$V(t) = V_0 e^{(- \frac{t}{\tau}) }$

where $V_0$ is the initial voltage, and $\tau$ is the time constant.

For our problem, we know

$V_0 = 185 \ V$

and

$V(10 \ s) = V_0 e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V$

So

$185 \ V \ e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V$

$e^{(- \frac{10 \ s}{\tau}) } = \frac{1.64 \ V}{ 185 \ V }$

$ln (e^{(- \frac{10 \ s}{\tau}) } ) = ln (\frac{1.64 \ V}{ 185 \ V })$

$- \frac{10 \ s}{\tau} = ln (\frac{1.64 \ V}{ 185 \ V })$

$\tau = \frac{-10 \s}{ln (\frac{1.64 \ V}{ 185 \ V }) }$

This gives us

$\tau = 2.1161 s$

and this is the time constant.

<h3>b.</h3>

At t = 18.8 s we got:

$V(18.8 \ s) = 185 \ V \ e^{(- \frac{18.8 \ s}{2.1161 s}) }$

$V(18.8 \ s) = 185 \ V \ e^{(- \frac{18.8 \ s}{2.1161 s}) }$

$V(18.8 \ s) = 0.0256 \ V$

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2 years ago