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saw5 [17]
2 years ago
6

If you were standing on top of a high mid-Pacific island at 15° north latitude, from which direction would you expect the wind t

o come? (Hint: don't forget to consider Coriolis' effect!)
Physics
1 answer:
Brrunno [24]2 years ago
8 0

Answer:

Northeast

Explanation:

For a person standing on a top high mid-pacific island on the 15° north latitude, the wind is expected to come from the northeast.

This is because of the Coriolis's effect. According to the Coriolis's effect the the objects moving in the northern hemisphere, tends to veer of their course to the clockwise direction. While moving clockwise from the north the east is witnessed. Thus, the wind is expected from the north east

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A physics student walks 100 meters in 80 seconds. The student stops for 30 seconds, and then walks 200 meters farther in 90 seco
Elan Coil [88]

Answer:

B i think is the answer

Explanation:

i feel like it is B because if you put them together and the answer is 1.5 so it is B

8 0
2 years ago
Based on the article “Will the real atomic model please stand up?,” why did J.J. Thomson experiment with cathode ray tubes? to s
PIT_PIT [208]

Answer:

B.) to determine that electric beams in cathode ray tubes were actually made of particles

Explanation:

This is the right answer i just took the quiz on edge.

3 0
2 years ago
a marine biologist wants to know the total vertical distance a dolphin travel during a jump with the surface of the water being
marin [14]

From the starting depth to the surface, the vertical distance is 35 ft.

From the surface to the peak of the jump, the vertical distance is 27 ft.

From the peak of the jump to the surface, the vertical distance is 27 ft.

From the surface to the ending depth, the vertical distance is 18 ft.

Then the total vertical distance is ...

  35 ft + 27 ft + 27 ft + 18 ft = 107 ft

3 0
2 years ago
Read 2 more answers
A dog travels north for 18 meters, east for 8 meters, South for 27 meters and then west for 8 meters. What is the distance the d
Afina-wow [57]
I'm really not sure if this is right but I'll try.
The distance that the dog traveled is probably all of the distances added up. I would guess that it's 67 meters in total. 
The displacement is a little more tricky but you pretty much have to put a mental map in your head. Since East and West are both 8 meters, they cancel each other out. He travels more southern and that means the displacement is 9 meters south of his original location
7 0
2 years ago
Suppose you first walk 12.0 m in a direction 20 owest of north and then 20.0 m in a direction 40.0osouth of west. How far are yo
alexgriva [62]

Answer:

R=19.5m

\theta = 4.65° S of W

Explanation:

Refer the attached fig.

displacement  of the x and y components

x-component displacement is (R_{x}) = A_{x}+B_{x}

= A \sin(20°) + B \cos(40°)

= -12.0\sin(20°) + 20.0\cos(40°)

= -19.425m

x-component displacement is (R_{y}) = A_{y}+ B_{y}

=  A \cos(20°) - B \sin(40°)

= 12.0\cos(20°) - 20.0\sin(40°)

= -1.579

resultant displacement

∴

R = \sqrt{R_{x}^{2} +R_{y}^{2} }  }

=\sqrt{(-19.425)^{2}+(-1.579)^{2}  }

=19.5m

\theta = \tan^{-1}\left | \frac{R_{x}}{R_{y}} \right |

\theta = \tan^{-1}\left | \frac{1.579}{19.425} \right |

\theta = 4.65° S of W

6 0
2 years ago
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