<u><em>PRIMARY </em></u>Waves Are Detected First Because They Move So Fast.
<u><em>RIGHT</em></u> Angles To The Direction of Movement.
A Kind Of Scale Used To Measure The Amount of Seismic Energy Released By An Earthquake <u><em>RICHTER SCALE</em></u>
Answer: 529.9 Hz
Explanation:
Here we need to use the Doppler equation, so we have:
f' = f*(v + v0)/(v - vs)
Here, f is the frequency = 500Hz
v is the velocity of the wave, = 334m/s
v0 is the velocity of the observer = 20m/s
vs is the velocity of the source = 0m/s
Then we have:
f' = 500Hz*(334m/s + 20m/s)/(334m/s) = 529.9 Hz
Answer:
Tangential speed or Rotational speed
Radius of nuclei is given by formula

now we can say volume of the nuclei is given as

now the density is given as
density = mass / volume
mass of nuclei = mass of neutron + mass of protons



Now density is given as

here we know that
= 1.2 fm


So from above we can say that density of all nuclei is almost same.
Q before connected = Q after connected C1V1+C2V2 = (C1+C2) V
C1= 3×10^-6 F
V1= 480v
C2= 4×10^-6 F
V2= 500v
(3×10^-6)×(480) + (4×10^-6)×(500) = (3×10^-6 + 4×10^-6) × V
Simplifying the above, we get:
( 1440× 10^-6) + (2000 ×10^-6) = (7 × 10^-6) × V.
Further simplified as:
3440 × 10^-6 = 7 × 10^-6 × V
Making V the subject
V = 491.43volts
Therefore the potential difference across each capacitor is 491.43v