Answer:
Maximum force, F = 1809.55 N
Explanation:
Given that,
Diameter of the anterior cruciate ligament, d = 4.8 mm
Radius, r = 2.4 mm
The tensile strength of the anterior cruciate ligament,
We need to find the maximum force that could be applied to anterior cruciate ligament. We know that the unit of tensile strength is Pa. It must be a type of pressure. So,
So, the maximum force that could be applied to anterior cruciate ligament is 1809.55 N
Answer:
Please see the given attachment.
Explanation:
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<h2><em><u>PLEASE</u></em><em><u> </u></em><em><u>MARK</u></em><em><u> </u></em><em><u>ME</u></em><em><u> </u></em><em><u>AS </u></em><em><u>BRAINLEST</u></em><em><u>.</u></em></h2>
Answer:
a) = 1.9 eV
= 3.04 10⁻¹⁹ J,b ) This means that some electrons are at the first excited level of the hydrogen atom, which is highly likely as the temperature rises.
Explanation:
a) To calculate the maximum kinetic energy of the expelled electrons let's use the relationships of the photoelectric effect
= h f - Φ
Where K is the kinetic energy, h the Planck constant that is worth 6.63 10⁻³⁴ Js, f the frequency and Φ the work function
The speed of light is related to wavelength and frequency
c = λ f
Let's analyze the work function, it is the energy needed to start an electron from a metal, in this case to start an electron from a hydrogen atom its fundamental energy is needed, so
Φ= E₀ = 13.6 eV
let's replace and calculate the energy of the incident photon
E = h c / λ
E = 6.63 10⁻³⁴ 3 10⁸/80 10⁻⁹
E = 2,486 10⁻¹⁸ J
Let's reduce to eV
E = 2,486 10⁻¹⁸ (1 eV / 1.6 10⁻¹⁹)
E = 15.5 eV
Now we can calculate the kinetic energy
= h c / f - fi
= 15.5 -13.6
= 1.9 eV
b) Extra energy = 10.2 eV
The total kinetic energy of electrons is
Total kinetic energy = 1.9 +10.2 = 12.1 eV
For the calculation we are assuming that all the electors are in the hydrogen base state, but for temperatures greater than 0K some electors may be in some excited state, so less energy is needed to tear them out of hydrogen atom.
Let's analyze this possibility
ΔE = E photon - Total kinetic energy electron
ΔE = 15.5 - 12.1
ΔE = 3.4 eV
If we use the Bohr ratio for the hydrogen atom
= 13.606 / n2
n = √ 13.606 / En
n = √ (13606 / 3.4)
n = 2
This means that some electrons are at the first excited level of the hydrogen atom, which is highly likely as the temperature rises.