Kinetic energy (KE) is calculated through the equation,
KE = 0.5 mv²
where m and v are mass and velocity, respectively. Substituting the known values from the given above,
KE = 0.5(50 kg)(5 m/s)² = 625 J
Thus, the kinetic energy of Jesse at the highest point is equal to 625 J.
your answer is going to be all of the above
Answer:
a) λ = 435 nm
, c) c) λ = 4052 nm, d) λ= 95 nm
Explanation:
A) To carry out this excitation, the energy of the laser must be greater than or equal to the energy of the transition of the hydrogen atom, whose states of energy are described by the Bohr model.
En = -13,606 / n² [eV]
therefore the energy of the transition is
ΔE = E₅ -E₂
ΔE = 13.606 (1 / n₂² - 1 / n₅²)
ΔE = 13.606 (1/2² - 1/5²)
ΔE = 2,85726 eV
now let's use Planck's equation
E = h f
the speed of light is related to wavelength and frequencies
c = λ f
f = c /λ
E = h c /λ
λ = h c / E
let's reduce the energy to the SI system
E = 2,85726 eV (1.6 10⁻¹⁹ J / 1 eV) = 4.5716 10⁻¹⁹ J
let's calculate
λ = 6,626 10⁻³⁴ 3 10⁸ / 4,5716 10⁻¹⁹
λ = 4.348 10⁺⁷ m (10⁹ nm / 1 m)
λ = 435 nm
B) photon emission processes from this state with n = 5 to the base state n = 1, can give transition
initial state n = 5
final state n = 4
ΔE = 13.606 (1/4² - 1/5²)
ΔE = 0.306 eV
λ = h c / E
λ = 4052 nm
n = 5
final ΔE (eV) λ (nm)
level
4 0.306 4052
3 0.9675 1281
2 2,857 435
1 13.06 95
n = 4
3 0.661 1876
2 2,551 486
1 11,905 104
n = 3
2 1.89 656
1 12.09 102.5
n = 2
1 10.20 121.6
c) λ = 4052 nm
d) λ= 95 nm
Answer:
18.15 m/s
Explanation:
Radius of the loop, R = 23 m.
Weight of the rider, w= 760 N.
Mass of the rider, m = w/g = 77.5 kg
Apparent weight at the top of the loop is 350 N.
At the top:
The Earth takes (365 and 1/4) days to orbit the sun. That's the length of our 'year'.
Since it's not really possible to make our calendar a quarter of a day longer every year, we make the calendar a <u><em>whole</em></u> day longer every 4 years.
So "Leap years" make up for the extra one-fourth day the Earth takes to orbit the sun.