While sitting in your car by the side of a country road, you are approached by your friend, who happens to be in an identical ca
1 answer:
Answer:
7.85m/sec
Explanation:
definition of terms
<em>speed of sound is 340m/s</em>
<em>sound is an example of wave source</em>
<em>speed is change in distance per time</em>
Perceived frequency of friend's horn = 266Hz.
beat=f1-f2
6=f1-260 Hz
f1=266Hz.
we make an assumption that the velocity of sound will be used, I will adopt 340m/sec.
=340m/sec.
(340 x 266/260)- 340 = 7.85m/sec
my friend will be approaching me at that speed
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We know that acceleration is change in velocity by time taken for that change.
In this case velocity change is 3.7 m/s
Time taken for this change = 60 ms =
So acceleration of frog =
= 61.66 m/
So acceleration of frog is 61.66 m/
o it is evident that frog is capable of remarkable accelerations.
Answer:
-40 kJ
80 kJ
Explanation:
Work is equal to the area under the pressure vs volume graph.
W = ∫ᵥ₁ᵛ² P dV
2.27) Pressure and volume are linearly related. When we graph P vs V, the area under the line is a trapezoid. So the work is:
W = ½ (P₁ + P₂) (V₂ − V₁)
W = ½ (100 kPa + 300 kPa) (0.1 m³ − 0.3 m³)
W = -40 kJ
2.29) Pressure and volume are inversely proportional:
pV = k
The initial pressure and volume are 500 kPa and 0.1 m³. So the constant is:
(500) (0.1) = k
k = 50
The final pressure is 100 kPa. So the final volume is:
(100) V = 50
V = 0.5
The work is therefore:
W = ∫ᵥ₁ᵛ² P dV
W = ∫₀₁⁰⁵ (50/V) dV
W = 50 ln(V) |₀₁⁰⁵
W = 50 (ln 0.5 − ln 0.1)
W ≈ 80 kJ
