By Boyle's law:
P₁V₁ = P₂V₂
70*8 = P<span>₂*4
</span>P<span>₂*4 = 70*8
</span>
P<span>₂ = 70*8/4 = 140
</span>
P<span>₂ = 140 kiloPascals.</span>
Determine the frequency and the speed of these waves. The wavelength is 8.6 meters and the period is 6.2 seconds. Now find speed using the v = f. λ equation<span>.</span>
Answer:
(a) 5142.86 m
(b) 317.5 m/s
(c) 49.3 degree C
Explanation:
m = 100 kg, Q = 1200 kcal = 1200 x 1000 x 4.2 = 504 x 10^4 J
(a) Let the altitude be h
Q = m x g x h
504 x 10^4 = 100 x 9.8 x h
h = 5142.86 m
(b) Let v be the speed
Q = 1/2 m v^2
504 x 10^4 = 1/2 x 100 x v^2
v = 317.5 m/s
(c) The temperature of normal human body, T1 = 37 degree C
Let the final temperature is T2.
Q = m x c x (T2 - T1)
504 x 10^4 = 100 x 4.1 x 1000 x (T2 - 37)
T2 = 49.3 degree C
D. 5.098 x 106 is the correct answer when reduced to the proper notation.