<span>The specific heat (or the amount of heat required to raise the temperature of a unit mass of a substance by 1 degree Celsius) of copper is about 0.386 J/g/degree Celsius. This means that if we supply 0.386 J of energy to 1 gram of copper, its temperature will increase by 1 degree Celsius.</span>
Answer:
a) 23.2 e V
b) energy of the original photon is 36.8 eV
Explanation:
given,
energy at ground level = -13.6 e V
energy at first exited state = - 3.4 e V
A photon of energy ionized from ground state and electron of energy K is released.
h ν₁ - 13.6 = K
K combine with photon in first exited state giving out photon of energy
= 26.6 e V
h c = 6.626 × 10⁻³⁴ × 3 × 10⁸ = 12400 e V A°
K + ( 3.4 ) = 26.6 e V
a) energy of free electron
K = 26.6 - 3.4 = 23.2 e V
b) energy of the original photon
h ν₁ - 13.6 = K
h ν₁ = 23.2 + 13.6
= 36.8 e V
energy of the original photon is 36.8 eV
The question here would be what is the volume of the room. The density of air that is given has no use. We simply multiply the dimensions given of the room to determine the volume.
<span>43.0m × 18.0m × 15.0m = 11610m^3 ( 3.28 ft / 1 m)^3 = 4.09 x 10^5 ft^3</span>
Answer:
Explanation:
<u>Charge of an Electron</u>
Since Robert Millikan determined the charge of a single electron is
Every possible charged particle must have a charge that is an exact multiple of that elemental charge. For example, if a particle has 5 electrons in excess, thus its charge is 
Let's test the possible charges listed in the question:
. We have just found it's a possible charge of a particle
. Since 3.2 is an exact multiple of 1.6, this is also a possible charge of the oil droplets
this is not a possible charge for an oil droplet since it's smaller than the charge of the electron, the smallest unit of charge
cannot be a possible charge for an oil droplet because they are not exact multiples of 1.6
Finally, the charge 
is four times the charge of the electron, so it is a possible value for the charge of an oil droplet
Summarizing, the following are the possible values for the charge of an oil droplet:
<h2>
Its velocity when it crosses the finish line is 117.65 m/s</h2>
Explanation:
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = ?
Time, t = 6.8 s
Displacement, s = 1/4 mi = 400 meters
Substituting
s = ut + 0.5 at²
400 = 0 x 6.8 + 0.5 x a x 6.8²
a = 17.30 m/s²
Now we have equation of motion v = u + at
Initial velocity, u = 0 m/s
Final velocity, v = ?
Time, t = 6.8 s
Acceleration, a = 17.30 m/s²
Substituting
v = u + at
v = 0 + 17.30 x 6.8
v = 117.65 m/s
Its velocity when it crosses the finish line is 117.65 m/s
