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4 years ago

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The length of a train is about 1,700 meters. If there are approximately 3.28 feet in one meter, what is the length of the train

in feet?
0.002 feet

557,600 feet

5,576 feet

518 feet

1 answer:

melisa1 [442]4 years ago

5 0

It should be 5,576 feet

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1. Two object accumulated a charge of

In-s [12.5K]

Answer:

The magnitude of the electric force between these two objects

will be: 181.274 N.

i.e. F $=181.274$ N

Step-by-step explanation:

As

Two object accumulated a charge of 4.5 μC and another a charge of 2.8 μC.

so

q₁ = 4.5 μC = 4.5 × 10⁻⁶ C

q₂ = 2.8 μC = 2.8 × 10⁻⁶ C

separated distance = d = 2.5 cm

Calculating the magnitude of the force between two charged objects using the formula:

$F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}$

$=\:\frac{4.5\times \:\:10^{-6}\:\times \:\:2.8\:\times \:\:10^{-6}}{4\:\times \:\left(3.14\right)\:\times \:\left(8.85\times \:10^{-12}\right)\times \left(2.5\times \:\:10^{-2}\right)^2}$

$=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:4\times \:8.85\pi \left(10^{-2}\times \:2.5\right)^2}$ ∵ $4.5\times \:10^{-6}\times \:2.8\times \:10^{-6}=10^{-12}\times \:12.6$

$=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:35.4\pi \left(10^{-2}\times \:2.5\right)^2}$ ∵ $\mathrm{Multiply\:the\:numbers:}\:4\times \:8.85=35.4$

$\mathrm{Cancel\:the\:common\:factor:}\:10^{-12}$

$=\frac{12.6}{35.4\pi \left(10^{-2}\times \:2.5\right)^2}$

$=\frac{12.6}{0.025^2\times \:35.4\pi }$ ∵ $\left(10^{-2}\times \:2.5\right)^2=0.025^2$

$=\frac{12.6}{0.022125\pi }$ ∵ $35.4\pi\times 0.025^2=0.022125\pi$

$=\frac{12.6}{0.06950 }$

F $=181.274$ N

Therefore, the magnitude of the electric force between these two objects will be: 181.274 N.

i.e. F $=181.274$ N

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