Answer:
final displacement lf = 0.39 m
Explanation:
from change in momentum equation:
given: m = 0.4kg, y/x = 19/85, li = 1.9 m,
\delta p = 1.27 kg*m/s.
putting all value to get the final displacement value
final displacement lf = 0.39 m
An ideal gas undergoes isothermal compression from an initial volume of 5.39 m3 to a final volume of 2.84 m3. There is 9.75 mol
1 answer:
Answer:
a) Work done by gas = -15.584 KJ
b) Energy transferred = 15.584 KJ
Explanation:
Given:
Initial volume of the gas, V₁ = 5.39 m³
Final volume, V₂ = 2.84 m³
Temperature, T = 27.2°C = 273+27.2 = 300.2K
Number of moles, n = 9.75 mol
a) Work done for the given isothermal process is
where, R is the ideal gas constant = 8.31 J/mol.K
substituting the values we get,
or
b) From the first law of thermodynamics
change in internal energy = Heat added - Work done
or,
Now for the isothermal process, ΔU = 0
thus,
or
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The electric field is zero at x = -16.45cm
Data;
- q1 = 3.4 μC
- q2 = -2.0 μC
- distance = 5cm
As the 3μC is larger than -2.0μC and the charges are opposite sign. The electric field will be zero at the negative axis.
Let the point be at x.
For an electric field to be equal to zero;
Let's solve for x using mathematical methods.
Solving the above quadratic equation;
The electric field is zero at x = -16.45cm
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Answer:
atomic concentration = 2 atoms/unit cell
lattice parameter: a= 3.22 x 10⁻¹⁰ m
atomic radius: r= 1.39 x 10⁻¹⁰m
Explanation:
The atomic concentration is the number of atoms that can fit into a unit cell. It is a known number for each unit cell crystal structure. For a BCC (body-centered cube) crystal structure, atomic concentration is 2 atoms/unit cell because there are a 1/8 part of an atom in each corner of the cube (1/8 x 8= 1 atom) and 1 central atom in the central position of the cube ⇒ n= 1 atom + 1 atom= 2 atoms/unit cell
In order to calculate the lattice parameter a, we introduce the atomic mass 95.94 g/mol and the density 10.22 g/cm³ in the expression for the volume of the cube:
Vc= a³=
a³= 3.12 x 10⁻²³ m³
⇒ a = ∛(3.12 x 10⁻²³ m³) = 3.22 x 10⁻¹⁰m
Once we know the lattice parameter a, we can calculate the atomic radius r by using the expression of a for a BCC structure:
a=
⇒ r= a x √3/4= (3.22 x 10⁻¹⁰ m) x √3/4 = 1.39 x 10⁻¹⁰ m