<em><u>One</u></em><em><u> </u></em><em><u>newton</u></em><em><u> </u></em><em><u>force</u></em><em> </em><em>is</em><em> </em><em>defined as t</em><em>h</em><em>e</em><em> </em><em>force</em><em> </em><em>that</em><em> </em><em>is</em><em> necessary to provide a mass of one kilogram with an acceleration of one metre per second per second. One newton is equal to a force of 100,000 dynes in the centimetre-gram-second (CGS) system, or a force of about 0.2248 pound </em><em>i</em><em>n</em><em> </em><em>the</em><em> </em><em>f</em><em>o</em><em>o</em><em>t</em><em>-</em><em>p</em><em>o</em><em>u</em><em>n</em><em>d</em><em>-</em><em> </em><em>s</em><em>e</em><em>c</em><em>o</em><em>n</em><em>d</em><em> </em><em>system</em><em>.</em>
The particles of the medium (slinky in this case) move up and down (choice #2) in a transverse wave scenario.
This is the defining characteristic of transverse waves, like particles on the surface of water while a wave travels on it, or like particles in a slack rope when someone sends a wave through by giving it a jolt.
The other kind of waves is longitudinal, where the particles of the medium move "left-and-right" along the direction of the wave propagation. In the case of the slinky, this would be achieved by giving a tensioned slinky an "inward" jolt. You would see that such a jolt would give rise to a longitudinal wave traveling along the length of the tensioned slinky. Another example of longitudinal waves are sound waves.
Answer:
v = 3.7 m/s
Explanation:
As the swing starts from rest, if we choose the lowest point of the trajectory to be the zero reference level for gravitational potential energy, and if we neglect air resistance, we can apply energy conservation as follows:
m. g. h = 1/2 m v²
The only unknown (let alone the speed) in the equation , is the height from which the swing is released.
At this point, the ropes make a 30⁰ angle with the vertical, so we can obtain the vertical length at this point as L cos 30⁰, appying simply cos definition.
As the height we are looking for is the difference respect from the vertical length L, we can simply write as follows:
h = L - Lcos 30⁰ = 5m -5m. 0.866 = 4.3 m
Replacing in the energy conservation equation, and solving for v, we get:
v = √2.g.(L-Lcos30⁰) = √2.9.8 m/s². 4.3 m =3.7 m/s
Absorbtion consists of when light strikes on an object and bounces off.
