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
The correct answer is: Angular velocity =
![1.208 * 10^{-9}](https://tex.z-dn.net/?f=1.208%20%2A%2010%5E%7B-9%7D%20)
rad/s
Explanation:
The angular velocity is given as:
ω =
![\frac{2\pi}{T}](https://tex.z-dn.net/?f=%20%5Cfrac%7B2%5Cpi%7D%7BT%7D%20)
--- (1)
Where T = 165 * (365 days) * (24 hours/day) * (60 minutes/hour) * (60 seconds/minute) = 5203440000 s
Plug in the value in (1):
ω =
![\frac{2\pi}{5203440000} = 1.208 * 10^{-9}](https://tex.z-dn.net/?f=%20%5Cfrac%7B2%5Cpi%7D%7B5203440000%7D%20%3D%201.208%20%2A%2010%5E%7B-9%7D%20)
rad/s
Explanation:
We know that the sky appears to us like a sphere called as celestial sphere which appears to rotate around an imaginary axis because of Earth's rotation. Since the axis cuts the celestial sphere at celestial poles all the object seems to circle around the celestial poles.
Condition 1: The stars rise and set perpendicular to the horizon
The observer is at the equator
Condition 2: The stars circle the sky parallel to the horizon
The observer is at the Pole of the Earth
Condition 3: The celestial equator passes through the zenith
The observer is at the equator
Condition 4: In the course of a year, all stars are visible
The observer is at the equator
Condition 5: The Sun rises on March 21 and does not set until September 21 (ideally)
The observer is at North Pole