A similar but separate notion is that of velocity, which the rate of change<span> of </span>position<span>. Example . If p(t) is the </span>position<span> of an </span>object<span> moving on a number line at time t (measured in minutes, say), then the average </span>rate of change<span> of p(t) is the average velocity of the </span>object<span>, measured in units per minute.</span>
Two analogies:
1. The wires are like hoses full of water. As soon as you turn on the water, water is pushed out the end of the hose.
2. Although the electrons only move at 1mm/s, the electomagnetic pulse travels through the circuit at near speed of light (varies btw .1c to .9c). It is this pulse that provides the pressure to push the electrons out the end of the wire into the light
I think the answer is 3, population.
In
order to determine the mass of a standard baseball if it had the same density
(mass per unit volume) as a proton or neutron, we first determine the volume of
the baseball. The formula to be used is V_sphere = (4/3)*pi*r^3. In this case, the
radius r can be obtained from the circumference C, C = 2*pi*r. After plugging
in C = 23 cm to the equation, we get r = 3.6066 cm. The volume of the baseball
is then equal to 205.4625 cm^3.
Next,
take note of these necessary information:
Mass of a neutron/proton
= 10^-27 kg
Diameter of a
neutron/proton = 10^-15 m
Radius of a
neutron/proton = [(10^-15)/2]*100 = 5x10^-14 cm
<span>Thus,
the density, M/V of the neutron/proton is equal to 1.9099x10^12 kg/cm^3. Finally,
the mass of the baseball if it was a neutron/proton can be determined by
multiplying the density of the neutron/proton with the volume of the baseball. The
final answer is then a large value of 3.9241x10^14 kg.</span>
Atom would be...hmmm let me think
