Answer:
11.78meters
Explanation:
Given data
Mass m = 100kg
Length of cord= 10m
Spring constant k= 35N/m
At the greatest vertical distance, the spring potential energy is equal to the gravitational potential energy
That is
Us=Ug
Us= 1/2kx^2
Ug= mgh
1/2kx^2= mgh
0.5*35*10^2= 100*9.81*h
0.5*35*100=981h
1750=981h
h= 1750/981
h= 1.78
Hence the bungee jumper will reach 1.78+10= 11.78meters below the surface of the bridge
Answer:
The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.
Explanation:
Given that,
The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.
The speed of sound in air is 343 m/s.
To find,
The wavelength range for the corresponding frequency.
Solution,
The speed of sound is given by the following relation as :

Wavelength for f = 45 Hz is,


Wavelength for f = 375 Hz is,


So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.