Solar energy heats the surface of the Earth, including the ground, rocks, and even roadways. As the temperature of these surfaces increases, heat energy is released back into the atmosphere, forming a pocket of warm air. As the warm air pocket rises, it is replaced by cooler air. This process is responsible for many of the weather patterns in our atmosphere and is known as
Answer:
convection
Explanation:
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Three charges lie along the x-axis. One positive charge, q1 = 4.80*10^-18 C, is at x = 3.72 m, and another positive charge, q2 = 1.60*10^-19 C, is at the origin.
At what point on the x-axis must a negative charge, q3, be placed so that the resultant force on it is zero?
Answer:
The third charge needs to be placed at [tex]x \approx 0.57\; \rm m[/tex].
Explanation:
Both [tex]q_1[/tex] and [tex]q_2[/tex] would attract [tex]q_3[/tex].
These two electrostatic attractions need to balance one another. Hence, they need to be opposite to one another. Therefore, [tex]q_1[/tex] and [tex]q_2[/tex] need to be on opposite sides of [tex]q_3[/tex]. That is possible only if [tex]q_3 \![/tex] is on the line segment between [tex]q_1 \![/tex] and [tex]q_2 \![/tex].
Assume that [tex]q_3[/tex] is at [tex]x\; \rm m[/tex], where [tex]0 < x < 3.72[/tex] (in other words, [tex]q_3 \![/tex] is on the line segment between [tex]q_1[/tex] and [tex]q_2[/tex], and is [tex]x\; \rm m \![/tex] away from [tex]q_2 \![/tex].)
Let [tex]k[/tex] denote Coulomb's constant.
The magnitude of the electrostatic attraction between [tex]q_1[/tex] and [tex]q_3[/tex] would be:
[tex]\displaystyle \frac{k\cdot q_1 \cdot q_3}{(3.72 - x)^{2}}[/tex].
Similarly, the magnitude of the electrostatic attraction between [tex]q_2[/tex] and [tex]q_3[/tex] would be:
[tex]\displaystyle \frac{k\cdot q_2 \cdot q_3}{x^{2}}[/tex].
The magnitudes of these two electrostatic attractions need to be equal to one another for the resultant electrostatic force on [tex]q_3[/tex] to be [tex]0[/tex]. Equate these two expressions and solve for [tex]x[/tex]:
[tex]\displaystyle \frac{k\cdot q_1 \cdot q_3}{(3.72 - x)^{2}} = \frac{k\cdot q_2 \cdot q_3}{x^{2}}[/tex].
[tex]\displaystyle \frac{q_1}{(3.72 - x)^{2}} = \frac{q_2}{x^{2}}[/tex].
[tex]\displaystyle \frac{x^2}{(3.72 - x)^{2}} = \frac{q_2}{q_1}[/tex].
[tex]\displaystyle \frac{x^2}{(3.72 - x)^{2}} = \frac{q_2}{q_1} = \frac{1}{30}[/tex].
By the assumption that [tex](0 < x < 3.72)[/tex], it should be true that [tex](x > 0)[/tex] and [tex](3.72 - x > 0)[/tex]. Therefore, [tex]\displaystyle \frac{x}{(3.72 - x)} > 0[/tex].
Take the square root of both sides of the equation [tex]\displaystyle \frac{x^2}{(3.72 - x)^{2}} = \frac{1}{30}[/tex].
[tex]\displaystyle \sqrt{\frac{x^2}{(3.72 - x)^{2}}} = \sqrt{\frac{1}{30}}[/tex].
[tex]\displaystyle \frac{x}{3.72 - x} = \frac{1}{\sqrt{30}}[/tex].
[tex]\sqrt{30}\, x = 3.72 - x[/tex].
Therefore:
[tex]\left(1 + \sqrt{30}\right)\, x = 3.72[/tex].
[tex]\displaystyle x = \frac{3.72}{1 + \sqrt{30}} \approx 0.57[/tex].
Hence, [tex]q_3[/tex] should be placed at [tex]x \approx 0.57\; \rm m[/tex].
Which of the following is an example of an electromagnetic wave ?
Answer:
Its a wave
Explanation:
THis is one
Answer: a light wave
Explanation: a pex
Explain how a current is induced in the wire?
Answer:
If a coil of wire is placed in a changing magnetic field, a current will be induced in the wire. This current flows because something is producing an electric field that forces the charges around the wire. (It cannot be the magnetic force since the charges are not initially moving). ... that determines the induced current.