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Akimi4 [234]
3 years ago
7

When a weather map shows a wind symbol at 270 it indicates that the wind is coming from the?

Physics
1 answer:
Montano1993 [528]3 years ago
5 0
For the answer to the question above, <span>When a weather map shows a wind symbol at 270 it indicates that the wind is coming from the WEST.
I hope my answer helped you. Feel free to ask more questions. Have a nice day!</span>
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A tennis racket strikes a tennis ball with a force of 100 N. With what force
viva [34]

Answer:

-100N

Explanation:

Newton's third law of motion states that to every force exerted on one body, there is an equal and opposite force. This means that if object A exerts an ACTION force on B, there is a force called REACTION FORCE, which is equal and opposite, exerted on A by B.

The action and reaction forces are equal in size/magnitude but opposite in direction. In this case where a tennis racket strikes a tennis ball with a force (action force) of 100N, the ball will strike the racket with a reaction force of -100N.

F(RB) = -F(BR)

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What kind of pollution causes nitrogen and phosphorous to enter bodies of water?
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Nutrient pollution. I just took the quiz and got it right
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Based on the map, which area has a scarcity of fresh water?
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Answer:

North Africa

Explanation:

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2 years ago
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Psychology is a<br><br>A) science <br><br>B) philosophy <br><br>C) theory <br><br>D) opinion
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3 years ago
While at the county fair, you decide to ride the Ferris wheel. Having eaten too many candy apples and elephant ears, you find th
Dovator [93]

Answer:

Explanation:

From the given information:

radius = 15 m

Time T = 23 s

a) Speed (v) = \dfrac{2 \pi r}{T}

v = \dfrac{2\times \pi \times 15}{23}

v = 4.10 m/s

b) The magnitude of the acceleration is:

a = \dfrac{v^2}{r} \\ \\ a = \dfrac{(4.10)^2}{15}

a = 1.12 m/s²

c) True weight = mg

Apparent weight = normal force

From the top;

the normal force = upward direction,

weight is downward as well as the acceleration.

true weight - normal force = ma  

apparent weight =mg - ma  

\dfrac{apparent \ weight}{true \  weight} = \dfrac{(mg - ma)}{(mg)}

=1- \dfrac{1.12}{9.8}

= 0.886 m/s²

d)

From the bottom;

acceleration is upward, so:

apparent weight - true weight = ma

apparent weight = true weight + ma

\dfrac{apparent \ weight }{true \ weight} =\dfrac{ mg + ma}{mg}

\dfrac{apparent \ weight }{true \ weight} =1+ \dfrac{a}{g} \\ \\ = 1 + \dfrac{1.12}{9.8}

= 1 + \dfrac{1.12}{9.8} \\ \\

= 1.114 m/s²

6 0
3 years ago
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