Answer:
Ф = 2.179 eV
Explanation:
This exercise has electrons ejected from a metal, which is why it is an exercise on the photoelectric effect, which is explained assuming the existence of energy quanta called photons that behave like particles.
E = K + Ф
the energy of the photons is given by the Planck relation
E = h f
we substitute
h f = K + Ф
Ф= hf - K
the speed of light is related to wavelength and frequency
c = λ f
f = c /λ
Φ =
let's reduce the energy to the SI system
K = 0.890 eV (1.6 10⁻¹⁹ J / 1eV) = 1.424 10⁻¹⁹ J
calculate
Ф = 6.63 10⁻³⁴ 3 10⁸/405 10⁻⁹ -1.424 10⁻¹⁹
Ф = 4.911 10⁻¹⁹ - 1.424 10⁻¹⁹
Ф = 3.4571 10⁻¹⁹ J
we reduce to eV
Ф = 3.4871 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)
Ф = 2.179 eV
Hi there!
We can use Newton's Second Law:

ΣF = Net force (N)
m = mass (kg)
a = acceleration (m/s²)
We can rearrange the equation to solve for the acceleration.

Answer:

Explanation:
-The chemical formula for Molybdenum (V) Dichromate is 
-There are 21 moles of oxygen per one mole of Molybdenum (V) Dichromate
-We apply Avogadro's constant to find the number of atoms of oxygen:

Hence, there are
A).
It would decrease because the speed of sound and temperature are proportional.
Answer:
21870.3156 N
Explanation:
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 1.6 m/s²
Equation of motion

The acceleration of the craft should be 1.02234 m/s²

Weight of the craft

Thrust

The thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface is 21870.3156 N