The correct choice is
D. 22 Hz and 42 Hz.
In fact, the beat frequency is given by the difference between the frequencies of the two waves:
In this problem, the beat frequency is 
, therefore the only pair of frequencies that gives a difference equal to 20 Hz is
D. 22 Hz and 42 Hz.
Answer:
18.03 N
Explanation:
From the fiqure below,
Using parallelogram law of vector
R² = 15²+5²-2×5×15cos(180-60)
R² = 225+25-150cos120°
R² = 250-150(-0.5)
R² = 250+75
R² = 325
R = √325
R = 18.03 N
Hence the resultant force of the object is 18.03 N
v = initial velocity of launch of the stone = 12 m/s
θ = angle of the velocity from the horizontal = 30
Consider the motion of the stone along the vertical direction taking upward direction as positive and down direction as negative.
v₀ = initial velocity along vertical direction = v Sinθ = 12 Sin30 = 6 m/s
a = acceleration of the stone = - 9.8 m/s²
t = time of travel = 4.8 s
Y = vertical displacement of stone = vertical height of the cliff = ?
using the kinematics equation
Y = v₀ t + (0.5) a t²
inserting the values
Y = 6 (4.8) + (0.5) (- 9.8) (4.8)²
Y = - 84.1 m
hence the height of the cliff comes out to be 84.1 m
Vf = Vo + at
Vf = 20 m/s
Vo = 50 m/s
a = ?
t = 15
Therefore
20 = 50 + 15a
20 - 50 = 15a
-30 = 15a
a = -30 / 15
a = -2 m/s²
