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2 years ago

USE INTERNET TO ANSWER THE FOLLOWIN 6. Research about the causes of the following: (use diagrams were necessary

Physics

1 answer:

Mrac [35]2 years ago

7 0

Answer:

Ocean currents are streams of water flowing constantly on the ocean surface in definite directions. The ocean currents may be warm or cold. The warm ocean currents originate near the equator and move towards the poles. The cold current carry water from polar or higher latitudes to tropical or lower latitudes. For example the Labrador Ocean current is a cold current while the Gulf Stream is a warm current. The ocean current influence the temperature conditions of the area. Warm currents bring about warm temperatures over land surface. The areas where the warm current and cold currents meet provide the best fishing grounds of the world. For example seas around Japan and the eastern coast of North America. The areas where a warm and cold current meet also experience foggy weather and therefore navigation becomes difficult.

In causes form

Climatic Conditions: Currents influence the climatic conditions of the regions in which they flow. The warm Equatorial currents raise the temperature of the region in which they flow. Similarly, the cold currents lower the temperature of the places where they flow.

For further information check the given link

below

<h3>https://www.conserve-energy-future.com/ocean-currents.php</h3>

<h3>https://byjus.com/free-ias-prep/ocean-currents/</h3>

<h3>https://www.pmfias.com/ocean-currents-factors-responsible-formation-ocean-currents-effects-ocean-currents/</h3>

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Tamiku [17]

Answer:

the angle is about 67.79 degrees

Explanation:

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We also know that the maximum height of the projectile is given by the square of its initial vertical component of the velocity (Vyi) divided by 2g, therefore half of such distance is :

$half\,\,max-height = \frac{v_{yi}^2}{4\,g}$

we can use this information to find the y component of the velocity at that height via the formula:

$v_{yf}^2-v_{yi}^2=-2\,g\,\Delta y\\\\v_{yf}^2-v_{yi}^2=-2\,g\,(\frac{v_{yi}^2}{4\,g} )\\v_{yf}^2=v_{yi}^2-\frac{v_{yi}^2}{2} \\v_{yf}^2=\frac{v_{yi}^2}{2}$

Now we use the information that tells us the speed of the projectile at this height to be 1 m/s. That should be the result of the vector addition of the vertical and horizontal components:

$1=\sqrt{v_{yf}^2+0.5^2} \\1=\sqrt{\frac{v_{yi}^2}{2} +0.5^2}\\1^2=\frac{v_{yi}^2}{2} +0.5^2}\\1-0.5^2=\frac{v_{yi}^2}{2} \\2(1-0.5^2)=v_{yi}^2\\1.5 = v_{yi}^2\\v_{yi}=\sqrt{1.5} \\$

Now we can use the arc-tangent to calculate the launching angle, since we know the two initial component of the velocity vector:

$tan(\theta)=\frac{v_{yi}}{v_{xi}} =\frac{\sqrt{1.5} }{0.5} \\\theta= arctan(\frac{\sqrt{1.5} }{0.5})=67.79^o$

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If F1 is the magnitude of the force exerted

aleksandrvk [35]

Answer: 3. F1 = F2

Explanation:

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