<h3><em>physical</em><em> </em><em>science</em><em> </em><em>deals</em><em> </em><em>with</em><em> </em><em>the</em><em> </em><em>study</em><em> </em><em>of</em><em> </em><em>physics</em><em> </em><em>chemistry</em></h3>
Explanation:
yan lng po Alam ko
Long time ago, people saw the constellations as patterns in the sky. They names these patterns and tell stories about them. What people saw laong time ago are just mere patterns which forms animals and shapes. We got the names of our constellations from the Greeks who named the constellations after the mythological heroes and mythological legends.
I think that by "Classical physics" is meant low speed things. By low speed, I think is meant speed far below very roughly half the speed of light, so that Relativistic, special or general, effects can be ignored. Or at least it is hoped that they can be ignored.
Fire extinguishers and rockets get propelled by forcing out large amounts of material (gases under very high pressure) through a nozzle, and the RECOIL from that propels something forward. So, if the action is the ejection of material, the reaction (recoil) is the ejector moving along the same line in the other direction. And that's an example of Newton's third law.
Given a propulsion system, the magnitude of the force recoiling on the ejector will change the momentum of the ejector, often written as the equation F=ma where F is the force, m is the mass being accelerated, and a being the acceleration.
Just as something will stay still until it is moved - inertia - so once set in uniform motion in a straight line, the thing will continue in that motion, theoretically for ever or until something alters its momentum. Newton's first law is to the effect of "every body continues in a state of rest or uniform motion in a straight line unless acted on by a resultant external force". Which, I think, is where the concept of inertia stems from.
I think that the above mostly tcuches on the 3 laws.Any more help needed, please ask.
We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m. So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY....... We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION: v² = u² + 2as 0 = u² - 2gh u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)
It is wavelength i think so!!!!