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2 years ago

Why are the significantly more thunderstorms in Florida than in California?

2 answers:

8 0

The answer to your question is Thunderstorms are a frequent part of Florida life. They occur in all seasons of the year in Florida, but they are more numerous during the warm season when the wind off the sea flows inland during the afternoon. On an annual basis, communities in Florida usually experience thunderstorms 75 to 105 days per year. In fact, Florida leads the United States annually in the number of thunderstorm days. Out of 100,000 thunderstorms that occur within the United States each year, approximately 1 out of every 10 storms can become severe, causing damage or posing a threat to life. The average thunderstorm contains approximately 275 million gallons of water, which is enough water to fill 416 Olympic-sized swimming pools! <span>HOPED I HELPED!</span>

Andreyy892 years ago

3 0

Because Florida is wet and humid, while California is dry and non-humid. Florida also contains lots of lakes which evaporate to create thunderstorms.

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A car traveled 176 miles on 6.4 gallons of fuel the car averaged how many miles per gallon

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2 years ago

Car A (mass 1100 kg) is stopped at a traffic light when it is rearended by car B(mass 1400 kg). Both cars then slide with locke

viva [34]

Answer:

Part a)

$v_a = 3.94 m/s$

Part b)

$v_b = 3.35 m/s$

Part C)

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

Explanation:

Part a)

As we know that car A moves by distance 6.1 m after collision under the frictional force

so the deceleration due to friction is given as

$a = -\frac{F_f}{m}$

$a = -\frac{\mu mg}{m}$

$a = - \mu g$

now we will have

$v_f^2 - v_i^2 = 2ad$

$0 - v_i^2 = 2(-\mu g)(6.1)$

$v_a = \sqrt{(2(0.13)(9.81)(6.1)}$

$v_a = 3.94 m/s$

Part b)

Similarly for car B the distance of stop is given as 4.4 m

so we will have

$v_b = \sqrt{2(0.13)(9.81)(4.4)}$

$v_b = 3.35 m/s$

Part C)

By momentum conservation we will have

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1400 v_b = 1100(3.94) + 1400(3.35)$

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

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