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irga5000 [103]
3 years ago
10

At its peak, a tornado is 53.0 m in diameter and carries 465-km/h winds. What is its angular velocity in revolutions per second?

Physics
1 answer:
expeople1 [14]3 years ago
8 0

Answer:

0.7757 rev/s

Explanation:

d = Diameter of the tornado = 53 m

r = Radius of the tornado = 53/2 = 26.5 m

v = Velocity of wind = 465 km/h

Converting velocity to m/s

465=465\times \frac{1000}{3600}=\frac{775}{6}

Angular velocity

\omega=\frac{v}{r}\\\Rightarrow \omega=\frac{\frac{775}{6}}{26.5}\\\Rightarrow \omega=4.87\ rad/s

\omega=4.87\frac{1}{2\pi}=0.7757\ rev/s

∴ Angular velocity is 0.7757 rev/s

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Simplifyng common terms, and solving for Tsat, we have:

Tsat = √((27.3)²/8) = 9.7 days

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3 years ago
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3 years ago
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Answer:

a) v = 88.54 m/s

b) vf = 26.4 m/s

Explanation:

Given that;

m = 1400.0 kg

a)

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mg × ( 3200-2800) = 1/2 ×m×v²

so

1400 × 9.8 × 400 = 0.5 × 1400 × v²

5488000 = 700v²

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we substitute

1400 × 9.8 × ( 3200-2800) - (5 × 10⁶) = 1/2 × 1400 × (vf²  -0 )

488000 = 700 vf²

vf² = 488000 / 700

vf² = 697.1428

vf = √697.1428

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Electrons have a negative charge.
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2 years ago
WILL GIVE BRAINLIST ​
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Answer:

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it shows that the cost of a tornado happeniy is 30 billion dollars

and the cost of an earthquake is way lower than the one the tornado which is 0 dollar

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2 years ago
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