Answer:
19.01 N
Explanation:
F = Force being applied to the crate = 45 N
= Angle at which the force is being applied = 
Horizontal component of force is given by

The horizontal component of the force acting on the crate is 19.01 N.
Air for a diver comes out of a high pressure tank at - Same- pressure compared to the water around the diver (metered by the regulator).
This means the lungs are inflated with - Highly pressurized- gas.
This does not adversely affect the diver when deep underwater, because the entire environment around the diver is at -Same - pressure.
If the diver suddenly surface, the air in the alveoli in the lungs will still be at - a higher - pressure compared to the air around the diver, which will be at - a lower - pressure.
The gas in the diver's lungs will - expand - and can damage the alveoli.
Because pH can unballence the are it is not good for are health
Answer:
Explanation:
Given:
volume of urine discharged, 
time taken for the discharge, 
diameter of cylindrical urethra, 
length of cylindrical urethra, 
density of urine, 
a)
we have volume flow rate Q:
& 
where:
cross-sectional area of urethra
velocity of flow




b)
The pressure required when the fluid is released at the same height as the bladder and that the fluid is at rest in the bladder:


