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Masja [62]
3 years ago
10

The diagram shows how air circulation forms a sea breeze on a warm day. A beach meets the sea in the middle of the image with a

blue sky above containing a bright sun. On top above the water is A pair of arrows changing from red to blue pointing right toward the sea labeled 1 Return Flow. Then a blue arrow points down over the sea labeled 2 Cool. Then a pair of arrows going from blue to red point left labeled 3 Sea Breeze. Then a red arrow points up over the sand labeled 4 Warm. Which best describes what is happening in the diagram? At point 1, the air in the convection current is warmer and denser than at point 3 because its molecules contain more kinetic energy. At point 2, the air absorbs thermal energy from the surrounding air, which causes it to become heavier and sink. At point 3, the air in the convection current is cooler and lighter than the air in other parts of the current, so it is preparing to rise. At point 4, thermal energy from warm air over the land is transferred upward toward cooler areas of the atmosphere
Physics
2 answers:
nadezda [96]3 years ago
7 0
Yes it was then a red left label
Mkey [24]3 years ago
3 0

Answer:

D

Explanation:

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A motorbike travels 100m in 2.5 seconds. What is its speed?
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Answer:

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Explanation:

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Part A If the velocity of a pitched ball has a magnitude of 47.5 m/s and the batted ball's velocity is 51.5 m/s in the opposite
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Explanation:

Let us assume that the mass of a pitched ball is 0.145 kg.

Initial velocity of the pitched ball, u = 47.5 m/s

Final speed of the ball, v = -51.5 m/s (in opposite direction)

We need to find the magnitude of the change in momentum of the ball and the impulse applied to it by the bat. The change in momentum of the ball is given by :

\Delta p=m(v-u)\\\\\Delta p=0.145\times ((-51.5)-47.5)\\\\\Delta p=-14.355\ kg-m/s

So, the magnitude of the change in momentum of the ball is 14.355 kg-m/s.

Let the the ball remains in contact with the bat for 2.00 ms. The impulse is given by :

J=\dfrac{\Delta p}{t}\\\\J=\dfrac{14.355}{2\times 10^{-3}}\\\\J=7177.5\ kg-m/s

Hence, this is the required solution.

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