When a warm air mass collides with a cold air mass, the warm air mass rises above the cold air mass leading too:
A) Sunny skies
B) Global warming
C) The formation of clouds and rain
D) Volcanic eruptions​

Answer:

convection

Explanation:

I hope this helps

Answer:

False

Explanation:

Neither the USGS nor any other scientists have ever predicted a major earthquake. USGS scientists can only calculate the probability that a significant earthquake will occur in a specific area within a certain number of years.

Answer:

Its a wave

Explanation:

THis is one

Answer: a light wave

Explanation: a pex

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.

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].

Answer:

The correct answer is: 0°C + 0°C = 32°F

Explanation:

We need to remember the conversion equation from Celsius to Fahrenheit:

[tex]y^{\circ}F=(x^{\circ}C * \frac{9}{5})+32[/tex]

In our case x = 0, then y will be:

[tex](0^{\circ}C * \frac{9}{5})+32=32[/tex]

[tex]y=32^{\circ}F[/tex]

Now 0°C + 0°C is just 0°C because if we add a body at a certain temperature to another body with the same temperature the total temperature will the same.

Then, knowing that 0°C = 32°F we can conclude that:

[tex]0^{\circ}C+0^{\circ}C=32^{\circ}F[/tex]

I hope it helps you!