When a warm air mass is trapped between two cooler air masses, it is called a/an
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Answer:
a. The volume V₁ and V₂
b. The case that involves the greatest momentum change = Case B
c. The case that involves the greatest impulse = Case B
d. b. The case that involves the greatest force = Case B
Explanation:
Here we have
Case A: V₁ = 150-mL, v₁ = 8 m/s
Case B: V₂ = 600-mL, v₁ = 8 m/s
a. The variable that is different for the two cases is the volume V₁ and V₂
b. The momentum change is by the following relation;
ΔM₁ = Mass, m × Δv₁
The mass of the balloon are;
Δv₁ = Change in velocity = Final velocity - Initial velocity
Mass = Density × Volume
Density of water = 0.997 g/mL
Case A, mass = 150 × 0.997 = 149.55 g
Case B, mass = 600 × 0.997 = 598.2 g
The momentum change is;
Case A: Mass, m × Δv₁ = 149.55 g/1000 × 8 m/s = 1.1964 g·m/s
Case B: Mass, m × Δv₁ = 598.2/1000 × 8 = 4.7856 g·m/s
Therefore Case B has the greatest momentum change
The case that has the gretest momentum change = Case B
c. The momentum change = impulse therefore Case B involves the greatest impulse
d. Here we have;
Impulse = Momentum change = × Δt = mΔV
∴ = m·ΔV/Δt
∴ For Case A = 149.55×8/Δt = 1196.4/Δt N
For Case B = 598.2×8/Δt = 4785.6/Δt
Where Δt is the same for Case A and Case B, for Case B >>
for Case B
Therefore, Case B involves the greatest force.
Answer:
f = 632 Hz
Explanation:
As we know that for destructive interference the path difference from two loud speakers must be equal to the odd multiple of half of the wavelength
here we know that
given that path difference from two loud speakers is given as
now we know that it will have fourth lowest frequency at which destructive interference will occurs
so here we have
now for frequency we know that