How can we reduce the ripple factor of a bridge recitifier from 0.48 to 0.1 approximately ?
1 answer:
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Answer:
The minimum thickness of the soap bubble for destructive interference to occur is 225.56 nm.
Explanation:
Given;
wavelength of light, λ = 600 nm
The minimum thickness of the soap bubble for destructive interference to occur is given by;
where;
n is refractive index of soap film = 1.33
Therefore, the minimum thickness of the soap bubble for destructive interference to occur is 225.56 nm.
Answer:
The final components of velocity of the composite object is 3.33 m/s.
Explanation:
Given;
mass of the first object, m₁ = 3.35 kg
initial velocity of the first object, u₁ = 4.90 m/s in positive x-direction
mass of the second object, m₂ = 1.88 kg
initial velocity of the second object, u₂ = 3.12 m/s in negative y-direction
initial momentum of the first object, P₁ = 3.35 x 4.9 = 16.415 kgm/s
initial momentum of the second object, P₂ = 1.88 x 3.12 = 5.8656 kgm/s
The resultant velocity of the two objects is given by;
R² = 16.415² + 5.8656²
R² = 303.858
R = √303.858
R = 17.432 kgm/s
Apply the principle of conservation of linear momentum for inelastic collision;
total initial momentum before = total final momentum after collision
P₁(x) + P₂(y) = Pf
R = Pf
R = v(m₁ + m₂)
17.432 = v(m₁ + m₂)
where;
v is the final components of velocity of the composite object
Therefore, the final components of velocity of the composite object is 3.33 m/s.
Answer:
the ship's energy is greater than this and the crew member does not meet the requirement
Explanation:
In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship
W =∫ F dx = ΔK
Let's replace
∫ (α x³ + β) dx = ΔK
α x⁴ / 4 + β x = ΔK
Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J
x (α x³ + β) = - K₀
= K₀ + x (α x³ + β)
Assuming that the low limit is x = 0, measured from the cargo hangar
Let's calculate
= 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)
Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)
Kf = 2.7 10¹¹ - 1.1475 10¹¹
Kf = 1.55 10¹¹ J
In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement
We evaluate the kinetic energy if the System is well calibrated
W = x F₀ = –K₀
= K₀ + x F₀
We calculate
= 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶
= (2.7 -2.625) 10¹¹
= 7.5 10⁹ J