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2 years ago

How does the Coriolis effect impact ocean currents in the Northern and Southern Hemispheres?

Physics

1 answer:

Strike441 [17]2 years ago

3 0

The Coriolis effect describes the pattern of deflection taken by objects not firmly connected to the ground as they travel long distances around the Earth.
The key to the Coriolis effect lies in Earth’s rotation. Specifically, Earth rotates faster at the Equator than it does at the poles. Earth is wider at the Equator, s

The Coriolis effect bends the direction of surface currents to the right in the Northern Hemisphere and left in the Southern Hemisphere.

Let’s pretend you’re standing at the Equator and you want to throw a ball to your friend in the middle of North America. If you throw the ball in a straight line, it will appear to land to the right of your friend because he’s moving slower and has not caught up.

Now let’s pretend you’re standing at the North Pole. When you throw the ball to your friend, it will again to appear to land to the right of him. But this time, it’s because he’s moving faster than you are and has moved ahead of the ball.

Everywhere you play global-scale "catch" in the Northern Hemisphere, the ball will deflect to the right.

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Show that the acceleration of any object down an incline where friction behaves simply (that is, where fk=μkN ) is a=g(sinθ−μkco

11Alexandr11 [23.1K]

Answer:

a=g(sinθ-μkcosθ)

Explanation:

In an inclined plane the forces that interact with the object can be seen in the figure. The normal force, the weight w and the decomposition of the force vector of weight can be observed.

wx=m*g*sinθ

wy=m*g*cosθ

As the objects moves down an incline, acceleration in y axis is 0.

Then, by second Newton's Law:

Fy = m*ay

FN - m*g cos θ = 0,

FN=m*g cos θ

In x axis the forces that interacs are the x component of weight and friction force:

Fx = m*ax

mg sen u-FN*μk=m*a

Being friction force, Fr=FN*μk, we replace with its value in below formula:

m*g *sinθ-(m*g*cosθ*μk)=m*a

Then, isolating a:

a=(m*g sinθ-(m*g*cosθ*μk))/m

Solving, we have next equation:

a=g sinθ-(g*cosθ*μk)

Applying distributive property we have:

a=g*(sinθ-μk*cosθ)

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What % of an object’s mass is above the water line if the object’s density is 0.82g/ml?

Sladkaya [172]

To develop this problem we will apply the Archimedes model. As well as the definitions of Weight based on mass and acceleration. The first in turn will be considered under the relationship of Density and Volume. From the values given we have to:

$\rho_w =1 g/mL \rightarrow \text{Water Density}$

$\rho_o = 0.82g/mL \rightarrow \text{Object density}$

Since it is in equilibrium, the weight of the object will have a reaction from the water, which will cause the sum of forces between the two objects to be zero, therefore

$\sum F= 0$