The work is path independent since we have a conservative force.
Thus
Answer (1)
Both organisms attempt to use the same limited sources
-- In combination with 610 Hz, the beat frequency is 4 Hz.
So the unknown frequency is either (610+4) = 614 Hz
or else (610-4) = 606 Hz.
In combination with 605 Hz, the beat frequency will be
either (614-605) = 9 Hz or else (606-605) = 1 Hz.
-- In actuality, when combined with the 605 Hz, the beat
frequency is too high to count accurately. That must be
the 9 Hz rather than the 1 Hz.
So the unknown is (605+9) = 614 Hz.
Answer:
just before landing the ground
Explanation:
Let the velocity of projection is u and the angle of projection is 30°.
Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.
initial horizontal component of velocity, ux = u Cos 30
initial vertical component of velocity, uy = u Sin 30
Time of flight is given by

Final horizontal component of velocity, vx = ux = u Cos 30
Let vy is teh final vertical component of velocity.
Use first equation of motion
vy = uy - gT


vy = - u Sin 30
The magnitude of final velocity is given by


v = u
Thus, the velocity is same as it just reaches the ground.
Answer:

Explanation:
from snells law we have following relation
is refractive index of air
n' - refractive index of air below
i is angle of incidence = 90 - 1.30 = 88.70^o
r is refraction angle
plugging all value in above formula

