I'm not sure about the distance to the nearest star, but it's probably about 4 light-years (L-y).
1 L-y = 1.86 * 10E5 mi/sec * 3600 sec/hr * 24 hr/day * 365 day/yr
1 L-y = 5.9 *10E12 mi and 4 L-y = 2.3 *10E13 mi distance to star
2.3 * 10E13 mi / 900 mi/hr = 2.6 * 10E10 hr hours to star
2.6 * 10E10 hr / (24 hr/day) = 1.1 * 10E9 day days to star
1.1 * 10E9 day / 365 day/yr = 3 * 10E6 yr = 3 million years to star
Answer:
a) E₀ = 2.125 eV, b) # photon2 = 9.2 10¹⁵ photons / mm²
Explanation:
a) To calculate the energy of a photon we use Planck's education
E = h f
And the ratio of the speed of light
c = λ f
We replace
E = h c /λ
Let's calculate
E₀ = 6.63 10⁻³⁴ 3 10⁸/585 10⁻⁹
E₀ = 3.40 10⁻¹⁹ J
Let's reduce
E₀ = 3.4 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)
E₀ = 2.125 eV
b) Let's look for the energy in each pulse
P = E / t
E = P t
E = 20.0 0.45 10⁻³
E = 9 10⁻³ J
let's use a ratio of proportions (rule of three) if we have the energy of a photon (E₀), to have the energy of 9 10⁻³ J
# photon = 9 10⁻³ /3.40 10⁻¹⁹
# photon = 2.65 10¹⁶ photons
Let's calculate the areas
Focus area
A₁ = π r²
A₁ = π (3.4/2)²
A₁ = 9,079 mm²2
Area requested for calculation r = 1 mm
A₂ = π 1²
A₂ = 3.1459 mm²
Let's use another rule of three. If we have 2.65 106 photons in an area A1 how many photons in an area A2
# photon2 = 2.65 10¹⁶ 3.1459 / 9.079
# photon2 = 9.2 10¹⁵ photons / mm²
Answer:
High specific heat -> takes more energy to raise/lower object's temperature
Low specific heat -> takes less energy to raise/lower object's temperature
Explanation:
The specific heat capacity is the amount of heat required to raise the temperature of something per unit of mass.
A high specific heat value for an object means it takes more energy to raise or lower that object's temperature. A low specific heat value for an object means it does not take very much energy to heat or cool that object.
Answer:
L = 1.545 m
Explanation:
Let the total length of the rod is L
now the torque must applied on the other end of the rod so that it will balance the torque due to weight of rock on other side of fulcrum
so we will have
so we have
F = 663 N
