Answer:
The water is flowing at the rate of 28.04 m/s.
Explanation:
Given;
Height of sea water, z₁ = 10.5 m
gauge pressure,
= 2.95 atm
Atmospheric pressure,
= 101325 Pa
To determine the speed of the water, apply Bernoulli's equation;

where;
P₁ = 
P₂ = 
v₁ = 0
z₂ = 0
Substitute in these values and the Bernoulli's equation will reduce to;

where;
is the density of seawater = 1030 kg/m³

Therefore, the water is flowing at the rate of 28.04 m/s.
Answer:
The phase difference between the reflected waves when they meet at the tuning fork is 159.29 rad.
Explanation:
Given that,
Frequency of sound wave = 240 Hz
Distance = 46.0 m
Distance of fork = 14 .0 m
We need to calculate the path difference
Using formula of path difference

Put the value into the formula


We need to calculate the wavelength
Using formula of wavelength

Put the value into the formula


We need to calculate the phase difference
Using formula of the phase difference

Put the value into the formula



Hence, The phase difference between the reflected waves when they meet at the tuning fork is 159.29 rad.
There are two atoms of potassium bonded to one atom of sulfur.
Use pythagorean theorem

to find the opposite side, which is 7.3
so then you can just use inverse sinA=7.3/10 which equals 46.9 degrees