Answer:
a)Amplitude ,A = 2 mm
b)f=95.49 Hz
c)V= 30 m/s ( + x direction )
d) λ = 0.31 m
e)Umax= 1.2 m/s
Explanation:
Given that
![y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]](https://tex.z-dn.net/?f=y%3D2%5C%20mm%5C%20sin%5B%2820m%5E%7B-1%7D%29x-%28600s%5E%7B-1%7D%29t%5D)
As we know that standard form of wave equation given as
A= Amplitude
ω=Frequency (rad /s)
t=Time
Φ = Phase difference
So from above equation we can say that
Amplitude ,A = 2 mm
Frequency ,ω= 600 rad/s (2πf=ω)
ω= 2πf
f= ω /2π
f= 300/π = 95.49 Hz
K= 20 rad/m
So velocity,V
V= ω /K
V= 600 /20 = 30 m/s ( + x direction )
V = f λ
30 = 95.49 x λ
λ = 0.31 m
We know that speed is the rate of displacement
![U=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]](https://tex.z-dn.net/?f=U%3D2%5C%20mm%5C%20sin%5B%2820m%5E%7B-1%7D%29x-%28600s%5E%7B-1%7D%29t%5D)
![U=1200\ cos[(20m^{-1})x-(600s^{-1})t]\ mm/s](https://tex.z-dn.net/?f=U%3D1200%5C%20cos%5B%2820m%5E%7B-1%7D%29x-%28600s%5E%7B-1%7D%29t%5D%5C%20mm%2Fs)
The maximum velocity
Umax = 1200 mm/s
Umax= 1.2 m/s
Answer:
f = 19,877 cm and P = 5D
Explanation:
This is a lens focal length exercise, which must be solved with the optical constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, p is the distance to the object and q is the distance to the image.
In this case the object is placed p = 25 cm from the eye, to be able to see it clearly the image must be at q = 97 cm from the eye
let's calculate
1 / f = 1/97 + 1/25
1 / f = 0.05
f = 19,877 cm
the power of a lens is defined by the inverse of the focal length in meters
P = 1 / f
P = 1 / 19,877 10-2
P = 5D
5.16 meters!!!!!!!!!!!!!!!!!!!!
Probably because of the drag coefficient and the density of the liquid.
The strong nuclear force holds the nucleus of an atom together.
Somehow, it overcomes the electrical force of repulsion between protons in the nucleus, which all have the same charge but still stay close together somehow. (b)
