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const2013 [10]
4 years ago
9

Which statement about tornadoes is correct? a. They are a very common part of thunderstorms. b. Seventy-five percent of the worl

d’s tornadoes occur in the United States. c. Their danger is mainly from the heavy rains they bring. d. They are the strongest storms on Earth.
Physics
2 answers:
Anit [1.1K]4 years ago
7 0

As strange as it may seem, around 75% world’s tornadoes

occur in the United States. It's actually because of the unique

geography of North America, and how the winds and storms

get funneled up through the US. (choice-b)


It's true that when there's a tornado, it usually comes along with a thunderstorm. But tornadoes are way way LESS common than thunderstorms.


Their main danger is from the high winds in the center of the funnel.

The rain is minor compared to the wind damage.


Tornadoes are not really the strongest storms on Earth.

Welll ... I guess that could depend on what "strong storm" means.

Sure, a tornado is incredibly vicious and powerful. But it's only a few

miles wide, and it only lasts for a few hours and then it's all over.

The tropical cyclones ... hurricanes, monsoons, typhoons ... don't have

winds as fast as tornadoes, but they can be a thousand miles wide and

last for weeks.

Likurg_2 [28]4 years ago
7 0
<h2><em><u>B</u></em></h2><h2><em><u>75% HAPPEN IN THE US</u></em></h2>
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A transverse wave on a string is described by the wave function
svp [43]

The period of the transverse wave from what we have here is 0.5

<h3>How to find the period of the transverse wave</h3>

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5 0
1 year ago
A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl
kondaur [170]

Answer:

Part a)

y_m = 0.157 mm

part b)

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

Explanation:

As we know that the speed of wave in string is given by

v = \sqrt{\frac{T}{m/L}}

so we have

T = 17.5 N

m/L = 5.4 g/cm = 0.54 kg/m

now we have

v = \sqrt{\frac{17.5}{0.54}}

v = 5.69 m/s

now we have

Part a)

y_m = amplitude of wave

y_m = 0.157 mm

part b)

k = \frac{\omega}{v}

here we know that

\omega = 2\pi f

\omega = 2\pi(92.2) = 579.3 rad/s

so we  have

k = \frac{579.3}{5.69}

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

4 0
3 years ago
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