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3 years ago

Most ocean waves obtain their energy and motion from _____. the moon's gravitational attraction the sun plate movement the wind

1 answer:

3 0

Hello!

Most ocean waves obtain their energy and motion from the wind.

Ocean waves are surface waves that move across the surface of the ocean. When wind touches the surface of the water, there is friction in the contact zone. This friction causes a drag effect, that makes wrinkles on the surface of the water. As the wrinkles get bigger, they transform into full-blown waves, and the taller the wave, the more energy it can extract from the wind, making them even bigger and to move longer distances.

Have a nice day!

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A 3,220 lb car enters an S-curve at A with a speed of 60 mi/hr with brakes applied to reduce the speed to 45 mi/hr at a uniform

Grace [21]

The magnitude of the total friction force exerted by the road on the tires at B will be 919.46 lb.

<h3>What is the friction force?</h3>

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. Its unit is Newton (N).

Mathematically, it is defined as the product of the coefficient of friction and normal reaction.

The given data in the problem is;

The weight is,W= 3,220 lb

The speed is,u= 60 mi/hr

The reducing speed is,v= 45 mi/hr

The distance traveled is,d= 300 ft

The radius of curvature of the path of the car at B is,R= 600 ft.

1 mile = 5280 ft

From the Newtons' equation of motion;

$\rm v^2 = u^2 +2ad \\\\ \rm (45 \times \frac{5280}{3600} )^2 = (60 \times \frac{5280}{3600} )^2 +2a\times 300 \\\\$

The tangential accelerations are;

$\rm a_t = \frac{66^2 -88^2}{600} \\\\ \rm a_t = -5.65 ft/sec^2 \\\\$

The force is found as;

$\rm \sum F = ma \\\\ \frac{3220}{32.2} \times -5.65 \\\\ F_T= 565 \ lb$

The normal force is;

$\rm F_n = \frac{3220}{32.2} \times \frac{66^2}{600} \\\\ F_N =726 \ lb$

The net or the total friction force exerted by the road on the tires at B. is found as;

$\rm F = \sqrt{(565)^2+(726)^2} \\\ F = 919.946 \ lb$

Hence, the magnitude of the total friction force exerted by the road on the tires at B will be 919.46 lb.

To learn more about the friction force, refer to the link;

brainly.com/question/1714663

#SPJ1

6 0

2 years ago

Suppose you have a total charge qtot that you can split in any manner. Once split, the separation distance is fixed. How do you

dybincka [34]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know from the Coulomb's Law that, Coulomb's force is directly proportional to the product of two charges q1 and q2 and inversely proportional to the square of the radius between them.

So,

F = $\frac{Kq1q2}{r^{2} }$

Now, we are asked to get the greatest force. So, in order to do that, product of the charges must be greatest because the force and product of charges are directly proportional.

Let's suppose, q1 = q

So,

if q1 = q

then

q2 = Q-q

Product of Charges = q1 x q2

Now, it is:

Product of Charges = q x (Q-q)

So,

Product of Charges = qQ - $q^{2}$

And the expression qQ - $q^{2}$ is clearly a quadratic expression. And clearly its roots are 0 and Q.

So, the highest value of the quadratic equation will be surely at mid-point between the two roots 0 and Q.

So, the midpoint is:

q = $\frac{Q + 0}{2}$

q = Q/2 and it is the highest value of each charge in order to get the greatest force.

8 0

3 years ago

Sophia is planning on going down an 8-m water slide. Her weight is 50 N. She knows that she has gravitational potential energy (

RideAnS [48]

Answer:

Explanation:

graph would be a straight line from (0, 0) to (400, 8)

Plot points are

PE = mgh

50(0) = 0 J

50(2) = 100 J

50(4) = 200 J

50(6) = 300 J

50(8) = 400 J

4 0

3 years ago

The object will feel minimum force of gravity at the

Alika [10]

Answer:

space , small amount of gravity is found in the space ,infact we can say that there is no gravity in the space

4 0

3 years ago

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An inventive child named Nick wants to reach an apple in a tree without climbing the tree. Sitting in a chair connected to a rop

Oduvanchick [21]

UHHH WHAT? I DONT GET THAT AT ALLOW

5 0

3 years ago