Answer:
Explanation:
A ) Distance between two adjacent anti-node will be equal to distance between two adjacent nodes . So the required distance is 15 cm .
B ) wave-length, amplitude, and speed of the two traveling waves that form this pattern are as follows
wave length = same as wave length of wave pattern formed. so it is 30 cm
amplitude = 1/2 the amplitude of wave pattern formed so it is .850 / 2 = .425 cm
Speed = frequency x wavelength ( frequency = 1 / time period )
= 1 / .075) x 30 cm
400 cm / m
C ) maximum speed
= ω A
= (2π / T) x A
= 2 X 3.14 x .85 / .075 cm / s
= 71.17 cm / s
minimum speed is zero.
D ) The shortest distance along the string between a node and an antinode
= Wavelength / 4
= 30 / 4
= 7.5 cm
The astrometric technique of planet detection works best for massive planets around nearby stars.
Answer:
Hi... Your answer is impulse
Answer:
λ = 5.65m
Explanation:
The Path Difference Condition is given as:
δ=
;
where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.
m = no of openings which is 2
∴δ= 
n is the index of refraction of the medium in which the wave is traveling
To find δ we have;
δ= 
δ= 
δ= 
δ= 
δ= 
δ= 
δ= 82.15 -73.68
δ= 8.47
Again remember; to calculate the wavelength of the ocean waves; we have:
δ=
δ= 8.47
8.47 =
λ = 
λ = 5.65m
