A tuning fork vibrating at 360 Hz falls from rest and accelerates at 9.8 m/s 2 .

Question

A tuning fork vibrating at 360 Hz falls from rest and accelerates at 9.8 m/s 2 . How far below the point of release is the tuning fork when waves of frequency 327.5 Hz reach the release point? The speed of sound in air is 339 m/s. Answer in units of m

b.

An unclothed student is in 20 ?C room.
If the skin temperature of the student is
37 ?C, how much heat is lost from his body
in 7 min, assuming that the emissivity of the
skin is 0.92 and the surface area of the student
is 1.5 m2. The Stephan-Boltzmann constant
is 5.6696

Explanation / Answer

step1:
fd=f(v+vr)/(v+vs) ,(fd=487Hz,f=511Hz,v=343m/s,vr=0,vs=?)
327.5= 360(339+0)/(339+vs)

vs= 29.64[m/s]

step2:
v=v0+at , (v=29.64m/s, v0=0m/s , a=9.8m/s^2 , t=?)

t= 3.02s

step3:
d=v0t+1/2at^2 , (v0=0, a=9.8m/s^2 , t= 3.028s , d=?)
d= 50.17[m] approx
-----------------------------------------------------------------------
T1 = 20 + 273 = 293 K
T2 = 37 + 273 = 310 K
e = 0.92
A = 1.5 m^2
?t = 7 x 60 = 420 s
? = 5.67 x 10^-8 W/m^2 x K^4

?Q/?t = e?A(T2^4 - T1^4)
?Q = (420s)(0.92)(1.5 m^2)(5.67 x 10^-8 W/m^2 x K^4)((310 K)^4 - (293 K)^4)
= 61295.32J (2dp)

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