let the vector Q is given as

given that

here we know that

now by above equation


so by comparing both sides
b = 0
4c - 3a = 6
also we know that


by solving above equation
a = 12.85 and c = 11.14
so the vector Q is given as

After the ball leaves the thrower's hand, the only force on it is
due to gravity. There's no horizontal force acting on it at all. (C)
Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance

Where, d = distance
A = amplitude
Put the value into the formula


Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
The answer is ultra violet radiation. From the air