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Greeley [361]
3 years ago
11

Is this statement true or false concerning squall line thunderstorm development? These often form ahead of the advancing front b

ut rarely behind it because lifting of warm, humid air and the generation of a squall line usually occur in the warm sector ahead of an advancing cold front. Behind a cold front, the air motions are usually downward, and the air is cooler and drier.
Physics
1 answer:
Pavlova-9 [17]3 years ago
3 0

Answer: The following statement is true about squall line thunderstorm development: <em><u>These often form ahead of the advancing front but rarely behind it because lifting of warm, humid air and the generation of a squall line usually occur in the warm sector ahead of an advancing cold front. Behind a cold front, the air motions are usually downward, and the air is cooler and drier.</u></em>

<em>An upper-level wave, accountable for the fabrication of a squall line, extend in front of and backside a cold front, the air backside the front is cold, steady and settling while the air ahead of the front is hot and co-seismic.</em>

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A van starts off 152 miles directly north from the city of Springfield. It travels due east at a speed of 25 miles per hour. Aft
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Answer:

12.84 miles per hour

Explanation:

Given:

Vertical distance of starting point of van from Springfield (d) = 152 miles

Speed in east direction (s) = 25 mph

Distance traveled in east direction (e) = 91 miles

Let the direct distance from Springfield of the van be 'x' at any time 't'.

Now, from the question, it is clear that, the vertical distance of van is fixed at 152 miles and only the horizontal distance is changing with time.

Now, consider a right angled triangle SNE representing the given situation.

Point S represents Springfield, N represents the starting point of van and E represents the position of van at any time 't'.

SN = d = 152 miles (fixed)

Now, using the pythagorean theorem, we have:

SE^2=SN^2+NE^2\\\\x^2=d^2+e^2\\\\x^2=(152)^2+e^2----(1)

Now, differentiating both sides with respect to time 't', we get:

2x\frac{dx}{dt}=0+2e\frac{de}{dt}\\\\\frac{dx}{dt}=\frac{e}{x}\frac{de}{dt}

Now, we are given speed as 25 mph. So, \frac{de}{dt}=25\ mph

Also, when e=91\ mi, we can find 'x' using equation (1). This gives,

x^2=23104+(91)^2\\\\x=\sqrt{31385}=177.16\ mi

Now, plug in the values of 'e' and 'x' and solve for \frac{dx}{dt}. This gives,

\frac{dx}{dt}=\frac{91}{177.16}\times 25\\\\\frac{dx}{dt}=12.84\ mph

Therefore, the distance between the van and Springfield is changing at a rate of 12.84 miles per hour

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