Answer:
Answer:
Speed of the wave in the string will be 3.2 m/sec
Explanation:
We have given frequency in the string fixed at both ends is 80 Hz
Distance between adjacent antipodes is 20 cm
We know that distance between two adjacent anti nodes is equal to half of the wavelength
So \frac{\lambda }{2}=20cm
2
λ
=20cm
\lambda =40cmλ=40cm
We have to find the speed of the wave in the string
Speed is equal to v=\lambda f=0.04\times 80=3.2m/secv=λf=0.04×80=3.2m/sec
So speed of the wave in the string will be 3.2 m/sec
Answer:
Generally, the lowest overtone for a pipe open at one end and closed would be at y / 4 where y represents lambda, the wavelength.
Since F (frequency) = c / y Speed/wavelength
F2 / F1 = y1 / y2 because c is the same in both cases
F2 = y1/y2 * F1
F2 = 3 F1 = 750 /sec
Note that L = y1 / 4 = 3 y2 / 4 for these wavelengths to fit in the pipe
and y1 = 3 y2
E=mgh. 196=5kg*9.81m/s^2*h. So h=196/(5*9.81)=4m
Answer:
v = 0
Explanation:
This problem can be solved by taking into account:
- The equation for the calculation of the period in a spring-masss system
( 1 )
- The equation for the velocity of a simple harmonic motion
( 2 )
where m is the mass of the block, k is the spring constant, A is the amplitude (in this case A = 14 cm) and v is the velocity of the block
Hence
and by reeplacing it in ( 2 ):
In this case for 0.9 s the velocity is zero, that is, the block is in a position with the max displacement from the equilibrium.
Reflecting telescope. Reflecting telescopes tend to have larger objective (due to the use of mirrors, mirrors are a lot cheaper than lenses) and have the ability to collect more light, while refracting telescopes are limited to objective lenses with smaller diameters due to their structural limitations (chromatic abbreviation, for example). Therefore, reflecting telescopes should be better at viewing faint distant stars
