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
![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)
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
<span>Nuclei of u-238 atoms are </span><span>unstable and spontaneously emit alpha particles.</span>
Answer:
liquid a particles slides past pother
Explanation:
mark brainliest :))
Answer:
From the data we know that runner A and runner B are 11 km apart from the start because (6+5) km
So the runner from the east direction has distance as unknown km, rate= 9 k/h ; time= d/r=x/9 hr
So runner towards the west will be
distance = 11-x, rate= 8 k/h, time = d/r = (11-x)/8
So equating east and west time we have
x/9= (11-x)/8
8x=99-9x
17x=99
x=5.92 km
That is the distance covered by runner towards the east and he will meet the runner toward the west at
6-5.92=0.08 km west of the flagpole.