The pyrite will be bigger, because its density is much lower.
I <em>do</em> know that the gold's volume will be 2.5906 (With a bunch more numbers after it)
50 divided by 19.3 = 2.5906
Answer:
Explanation:
The energy of a photon is given by the equation
, where h is the <em>Planck constant</em> and f the frequency of the photon. Thus, N photons of frequency f will give an energy of
.
We also know that frequency and wavelength are related by
, so we have
, where c is the <em>speed of light</em>.
We will want the number of photons, so we can write

We need to know then how much energy do we have to calculate N. The equation of power is
, so for the power we have and considering 1 second we can calculate the total energy, and then only consider the 4% of it which will produce light, or better said, the N photons, which means it will be
.
Putting this paragraph in equations:
.
And then we can substitute everything in our equation for number of photons, in S.I. and getting the values of constants from tables:

Thermal energy, radiant energy
Answer:
M g H / 2 = M g L / 2 initial potential energy of rod
I ω^2 / 2 = 1/3 M L^2 * ω^2 / 2 kinetic energy attained by rod
M g L / 2 = 1/3 M L^2 * ω^2 / 2
g = 3 L ω^2
ω = (g / (3 L))^1/2
Answer:
The actual angle is 30°
Explanation:
<h2>Equation of projectile:</h2><h2>y axis:</h2>

the velocity is Zero when the projectile reach in the maximum altitude:

When the time is vo/g the projectile are in the middle of the range.
<h2>x axis:</h2>

R=Range


**sin(2A)=2sin(A)cos(A)
<h2>The maximum range occurs when A=45°
(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>
Let B the actual angle of projectile

2B=60°
B=30°