Answer:
Explanation:
Inductance = 250 mH = 250 / 1000 = 0.25 H
capacitance = 4.40 µF = 4.4 × 10⁻⁶ F ( µ = 10⁻⁶)
ΔVmax = 240, f frequency = 50Hz and I max = 110 mA = 110 /1000 = 0.11A
a) inductive reactance = 2πfl = 2 × 3.142 × 50 × 0.25 H =78.55 ohms
b) capacitive reactance = = 1 / ( 2 × 3.142× 50 × 4.4 × 10⁻⁶ ) = 723.34 ohms
c) impedance = = 240 / 0.11 = 2181.82 ohms
The thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.
<h3>Thickness of the aluminum</h3>
The thickness of the aluminum can be determined using from distance of closest approach of the particle.
where;
Since the magnitude of charge of electron and proton is the same, at equal kinetic energy, the thickness will be same. r = 1.5 x 10⁻¹⁴ m.
<h3>For 10 MeV alpha-particles</h3>Charge of alpah particle = 2e
Thus, the thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.
Learn more about closest distance of approach here: brainly.com/question/6426420
Answer:
the only one that meets the requirements is option C .
Explanation:
The tolerance of a quantity is the maximum limit of variation allowed for that quantity.
To find it we must have the value of the magnitude, its closest value is the average value, this value can be given or if it is not known it is calculated with the formula
x_average = ∑ / n
The tolerance or error is the current value over the mean value per 100
Δx₁ = x₁ / x_average
tolerance = | 100 -Δx₁ 100 |
bars indicate absolute value
let's look for these values for each case
a)
x_average = (2.1700000+ 2.258571429) / 2
x_average = 2.2142857145
fluctuation for x₁
Δx₁ = 2.17000 / 2.2142857145
Tolerance = 100 - 97.999999991
Tolerance = 2.000000001%
fluctuation x₂
Δx₂ = 2.258571429 / 2.2142857145
Δx2 = 1.02
tolerance = 100 - 102.000000009
tolerance 2.000000001%
b)
x_average = (2.2 + 2.29) / 2
x_average = 2,245
fluctuation x₁
Δx₁ = 2.2 / 2.245
Δx₁ = 0.9799554
tolerance = 100 - 97,999
Tolerance = 2.00446%
fluctuation x₂
Δx₂ = 2.29 / 2.245
Δx₂ = 1.0200445
Tolerance = 2.00445%
c)
x_average = (2.211445 +2.3) / 2
x_average = 2.2557225
Δx₁ = 2.211445 / 2.2557225 = 0.9803710
tolerance = 100 - 98.0371
tolerance = 1.96%
Δx₂ = 2.3 / 2.2557225 = 1.024624
tolerance = 100 -101.962896
tolerance = 1.96%
d)
x_average = (2.20144927 + 2.29130435) / 2
x_average = 2.24637681
Δx₁ = 2.20144927 / 2.24637681 = 0.98000043
tolerance = 100 - 98.000043
tolerance = 2.000002%
Δx₂ = 2.29130435 / 2.24637681 = 1.0200000017
tolerance = 2.0000002%
e)
x_average = (2 +2,3) / 2
x_average = 2.15
Δx₁ = 2 / 2.15 = 0.93023
tolerance = 100 -93.023
tolerance = 6.98%
Δx₂ = 2.3 / 2.15 = 1.0698
tolerance = 6.97%
Let's analyze these results, the result E is clearly not in the requested tolerance range, the other values may be within the desired tolerance range depending on the required precision, for the high precision of this exercise the only one that meets the requirements is option C .