The separation in time between the arrival of primary and secondary wave is called LAG TIME.
The time difference between the arrival of primary wave and secondary wave in a seismogram is called lag time. The primary wave always travels faster than the secondary wave, thus the difference between the two can be obtained by estimating the difference between the arrival time of the two waves/.
Answer:
our sun is one of the least 100 billion stars in the milky way a spiral galaxy about 100000 light years across
Answer:
r = 0.5 m
Explanation:
First we find the angular speed of the ball by using its period:
ω = θ/t
For the time period:
ω = angular speed = ?
θ = angular displacement = 2π rad
t = time period = 0.5 s
Therefore,
ω = 2π rad/0.5 s
ω = 12.56 rad/s
Now, for the radius:
v = rω
r = v/ω
where,
v = linear speed = 6.29 m/s
r = radius = ?
r = (6.29 m/s)/(12.56 rad/s)
<u>r = 0.5 m</u>
0.23 mm far apart are the second-order fringes for these two wavelengths on a screen 1.5 m away.
<h3>Given wavelengths 710nm and 660nm,0.65mm apart two slits, and a screen 1.5m away.</h3>
Position of n the order fringe = n λ D / d
for n = 2
position = 2 λ D / d
λ = 710 nm , D = 1.5m
d = .65 x 10⁻³
position 1 = 2 x 710 x 10⁻⁹ x 1.5 / .65 x 10⁻³
= 3276.92 x 10⁻⁶ m
= 3.276x 10⁻³ m
= 3.276mm .
For λ = 660 nm
position = 2 λ D / d
λ = 660 nm , D = 1.5 m
d = .65 x 10⁻³
position 2 = 2 x 660 x 10⁻⁹ x 1.5 / .65 x 10⁻³
= 3046.15 x 10⁻⁶ m
= 3.046 x 10⁻³ m
= 3.046 mm .
Difference between their position
= 3.276mm ₋ 3.046 mm
= 0.23 mm .
To know more about Fringes refer to: brainly.com/question/15649748
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