Answer with Explanation:
We are given that
Fundamental frequency,f=88.4 Hz
Speed of sound,v=343 m/s
We have to find the first three overtones.
Tube is closed
The overtone of closed pipe is equal to odd number of fundamental frequency.
Therefore, the overtone of tuba
Where n=3,5,7,..
Substitute n=3
For second overtone
For third overtone
Answer:
11109.825 N
Explanation:
Given Data:
total mass =m=1510 kg
initial acceleration (a) =0.75g ( g=9.81 m/s² )
F=ma
= (1510)*( 0.75*9.81)
= 11109.825 N
Solution
distance travelled by Chris
\Delta t=\frac{1}{3600}hr.
X_{c}= [(\frac{21+0}{2})+(\frac{33+21}{2})+(\frac{55+47}{2})+(\frac{63+55}{2})+(\frac{70+63}{2})+(\frac{76+70}{2})+(\frac{82+76}{2})+(\frac{87+82}{2})+(\frac{91+87}{2})]\times\frac{1}{3600}
=\frac{579.5}{3600}=0.161miles
Kelly,
\Delta t=\frac{1}{3600}hr.
X_{k}=[(\frac{24+0}{2})+(\frac{3+24}{2})+(\frac{55+39}{2})+(\frac{62+55}{2})+(\frac{71+62}{2})+(\frac{79+71}{2})+(\frac{85+79}{2})+(\frac{85+92}{2})+(\frac{99+92}{2})+(\frac{103+99}{2})]\times\frac{1}{3600}
=\frac{657.5}{3600}
\Delta X=X_{k}-X_{C}=0.021miles
