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:
Vf=3
Explanation:
you must first write your data
data before impact
M1=1000 M2=5000
V1=0 m/s V2 =0m/s
data after impact
M1=1000 M2=5000
V1=15m/s V2=?
M1V1 +M2V2=M1V1 +M2V2f
(1000)(0)+(5000)(0)=(1000)(15)+(5000)Vf
0=15000+5000Vf
- 15000÷5000=5000Vf÷5000
Vf= -3
Vf =3
Answer: Gamma rays, x-rays, ultraviolet rays, visible light, and infrared rays.