if a water wave vibrates up and down 4 times each second,the distance between 2 successive crest is 5 meters, and the height fro
2 answers:
Answer:
4 Hz
5 m
1 m
20 m/s
Explanation:
The first part of the sentence describes the frequency of the wave i.e., the number of oscillations every second
f = 4 /s = 4 Hz
The second part of the sentence describes the wavelength or period of the wave i.e., the distance between successive crests or troughs.
λ = 5 m
The last part of the sentence describes the peak to peak amplitude of the wave i.e., the height between the lowest part to highest part of the wave
The peak amplitude is given by
The velocity of a wave is given by
v = fλ
⇒v = 4×5
⇒v = 20 m/s
The velocity of the wave is 20 m/s
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Answer:
Explanation:
Energy of signal being radiated per second on all sides = 71 x 10³ J .
At a distance of 220 m it is spread over an area of 4 π x (220)² because it is spreading uniformly on all sides.
So energy crossing per unit area
=
= 11.67 x 10⁻² Wm⁻²s⁻¹.
This is the intensity of the signal.
At 2200 m this intensity will further reduce by 100 times
So there it becomes equal to
11.67 x 10⁻⁴ Wm⁻² s⁻¹.
1) The electric potential at a distance r from a single point charge is given by
where k is the Coulomb's constant, q is the charge and r is the distance from the charge.
The charge in this problem is
So the potential at distance
is
2) By using the same formula as before, we can find the electric potential at distance r=99 m from the charge:
where k is the Coulomb's constant, q is the charge and r is the distance from the charge.
The charge in this problem is
So the potential at distance
2) By using the same formula as before, we can find the electric potential at distance r=99 m from the charge: