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GaryK [48]
2 years ago
8

How far will a car travel if its velocity is 15 m/s in 3 seconds? Follow example below.

Physics
1 answer:
svet-max [94.6K]2 years ago
4 0
<h3><u>Solution</u><u>:</u></h3>
  • Distance (d) = 112 m
  • Time (t) = 4 seconds
  • Let the speed be v.
  • We know, speed = Distance / Time
  • Therefore, v = d/t

or, v = 112 m ÷ 4 s = 28 m/s

<h3><u>Answer</u><u>:</u></h3>

<u>The </u><u>speed </u><u>of </u><u>the</u><u> </u><u>cheetah</u><u> </u><u>is </u><u>2</u><u>8</u><u> </u><u>m/</u><u>s.</u>

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Two equal charges with magnitude Q and Q experience a force of 12.3442 when held at a distance r. What is the force between two
andre [41]

Answer:

197.5072.

Explanation:

According to the Coulomb's law, the magnitude of the electrostatic force of interaction between two charges \rm q_1 and \rm q_2 which are separated by the distance \rm d is given by

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<em>where,</em> k is the Coulomb's constant.

For the case, when,

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Then, using Coulomb's law,

\rm 12.3442 = \dfrac{kQQ}{r^2}=\dfrac{kQ^2}{r^2}\ \ \ \ .......\ (1).

For the case, when,

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  • \rm q_2 = 2Q.
  • \rm d=\dfrac r2.

Then, using Coulomb's law, the new electric force between the charges is given by,

\rm F' = \dfrac{k(2Q)(2Q)}{\left (\dfrac r2\right )^2}\\=\dfrac{k\ 4Q^2}{\dfrac{r^2}{4}}\\=4\times 4 \times \dfrac{kQ^2}{r^2}\\=16\ \dfrac{kQ^2}{r^2}\\=16\times 12.3442\ \ \ \ \ \ \ \ (Using\ (1))\\=197.5072.

8 0
2 years ago
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2 years ago
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