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1 year ago

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Points A, B, and C form the vertices of a triangle in a nonuniform electrostatic field. The electrostatic work done on a particl

e of charge q as the particle travels from A to B is WAB and that done on the particle as it travels from A to C is WAC=−WAB/3.
a) How much electrostatic work is done on the particle as it travels from B to C?

1 answer:

Trava [24]1 year ago

4 0

Answer:

Explanation:

Let electric potential at A ,B and C be Va , Vb and Vc respectively.

Work done = charge x potential difference

Wab = q ( Va - Vb )

Wac = q ( Va - Vc )

Given

Wac = - Wab / 3

3Wac = - Wab

Now

Wbc = q ( Vb - Vc )

= q [ ( Va-Vc ) - ( Va - Vb )]

= Wac - Wab

= Wac + 3Wac

= 4Wac

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x = 1,185 m , t = 4/3 s , F = - 4 N

Explanation:

For this exercise we use Newton's second law

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dv = (β – α t) dt

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We evaluate between the lower limits v = v₀ for t = 0 and the upper limit v = v for t = t

v-v₀ = β t - ½ α t²

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t = 2 4/6

t = 4/3 s

Let's find the distance at this time

v = dx / dt

dx / dt = v₀ + β t - ½ α t2

dx = (v₀ + β t - ½ α t2) dt

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x = v₀ t + ½ β t - ½ 1/3 α t³

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1 year ago

When a resistor with resistance R is connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical powe

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4.41 W

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P = IV, V = IR

P = V² / R

Given that P = 0.0625 when V = 1.50:

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So the resistor is 36Ω.

When the voltage is 12.6, the power consumption is:

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1 year ago

In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. P

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Answer:

Time period for first satellites 24.46 days and for second satellites 37.67 days

Explanation:

Given :

Distance of first satellites $r_{sat1} = 48000 \times 10^{3}$ m

Distance of second satellites $r _{sat2} = 64000 \times 10^{3}$ m

Distance of charon $r_{c} = 19600 \times 10^{3}$ m

Time period of charon $T_{c} = 6.39$ days

From the kepler's third law,

Square of the time period is proportional to the cube of the semi major axis.

$T^{2} = r^{3}$

$\frac{T}{r^{\frac{3}{2} } } = constant$

For first satellites,

$\frac{T_{c} }{r_{c} ^{\frac{3}{2} } } = \frac{T_{sat1} }{r_{sat1} ^{\frac{3}{2} } }$

${T_{sat1} } = 6.39 \times \frac{(48000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}$

$T_{sat1} = 24.46$ days

For second satellites,

$\frac{T_{c} }{r_{c} ^{\frac{3}{2} } } = \frac{T_{sat2} }{r_{sat2} ^{\frac{3}{2} } }$

${T_{sat2} } = 6.39 \times \frac{(64000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}$

$T_{sat2} = 37.67$ days

Therefore, time period for first satellites = 24.46 days and for second satellites 37.67 days

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2 years ago

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Answer:

$\frac{E_1}{E_2}= 4$

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Capacitance C is given by

$C= \frac{\epsilon_0A}{d}$

A= area of capacitor cross section

d= distance

therefore,

$C_1= \frac{\epsilon_0A_1}{d_1}$

A_1= πR^2

d_1= d

$C_2= \frac{\epsilon_0A_2}{d_2}$

A_= π(2R)^2

d_2 = 2d

$q= \frac{\epsilon_0A_1}{d_1}V_1$

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$V_1= \frac{qd_1}{\epsilon_0A_1}$

and

$V_2= \frac{qd_2}{\epsilon_0A_2}$

also we know that E= V/d

⇒ $\frac{E_1}{E_2}= \frac{V_1}{V_2}\times\frac{d_2}{d_1}$

⇒ $\frac{E_1}{E_2}= \frac{qd_1}{\epsilon_0A_1}\times\frac{\epsilon_0A_2}{qd_2}\times\frac{d_2}{d_1}$

= A_1/A_2= $\frac{4R^2}{R^2}$ =4

therefore,

$\frac{E_1}{E_2}= 4$

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1 year ago