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saul85 [17]
2 years ago
11

Looking down on a Northern Hemisphere extratropical cyclone, surface winds blow ________ about the center. a.counterclockwise an

d inward b.clockwise and outward
Physics
1 answer:
Lesechka [4]2 years ago
4 0

Answer:

counterclockwise and inward

Explanation:

Extratropical cyclones or mid-latitude cyclones are the low-pressure areas, along with anticyclones of the high-pressure areas that drive weather over much of the Earth.

<u>Due to the Coriolis effect, the flow of the wind around the extratropical cyclone is counterclockwise in direction in the northern hemisphere, and clockwise in direction in the southern hemisphere.</u>

<u>It forms a comma shape which drives the forces inward the center.</u>

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During one trial, the acceleration is 2m/s^2 to the right. What calculation will give the tensions in actin filaments during thi
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Answer: hello your question is poorly written attached below is the complete question

answer :

TA = 1.6*10^-24 * 60 * 2,  TB = 1.6*10^-24 * ( 60 + 30 ) * 2  -- ( option 1 )

Explanation:

a = 2m/s^2

Ta = m₁ a = 60 * 1.6 * 10^-24 * 2 ц

Tb - Ta = m₂ a

∴ Tb = m₂ a  + Ta

       = ( 30 * 1.6 * 10^-24 * 2 ) +  ( 60 * 1.6 * 10^-24 * 2 )

= ( 30 + 60 ) * 1.6 * 10^-24 * 2 ц

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3 years ago
Volume of water shrinks between__​
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Answer:

When liquid water is cooled, it contracts like one would expect until a temperature of approximately 4 degrees Celsius is reached. After that, it expands slightly until it reaches the freezing point, and then when it freezes it expands by approximately 9%.

5 0
3 years ago
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A hollow conducting spherical shell has radii of 0.80 m and 1.20 m, The radial component of the electric field at a point that i
mars1129 [50]

Complete Question

The complete question is  shown on the first uploaded image  

 

Answer:

The electric field at that point is  E = 7500 \ N/C

Explanation:

From the question we are told that  

       The  radius of the inner circle is r_i  =  0.80  \ m

        The  radius of the outer circle is  r_o  =  1.20 \ m

       The  charge on the spherical shell q_n  =  -500nC  = -500*10^{-9} \ C

      The magnitude of the point charge at the center is  q_c =  + 300 nC  =  + 300 * 10^{-9} \ C

        The  position we are considering is  x =  0.60 m  from the center

Generally  the  electric field  at the distance x =  0.60 m  from the center  is mathematically represented as

                 E =  \frac{k *  q_c   }{x^2}

substituting values  

                  E =  \frac{k *  q_c   }{x^2}

where  k is  the coulomb constant with value k = 9*10^{9}  \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.

     substituting values

                  E =  \frac{9*10^9  *  300 *10^{-9}}{0.6^2}

                 E = 7500 \ N/C

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Answer:

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I hope this helps a little bit

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