Answer:
442.36038 m or 1451.31362 ft
Explanation:
= Initial pressure = 30.15 inHg
= Final pressure = 28.607 inHg
= Density of air = 0.075 lb/ft³



Density of mercury = 13560 kg/m³
g = Acceleration due to gravity = 9.81 m/s²
Difference in pressure is given by

The height of the building is 442.36038 m or 1451.31362 ft
Answer:
The first part can be solved via conservation of energy.

For the second part,
the free body diagram of the car should be as follows:
- weight in the downwards direction
- normal force of the track to the car in the downwards direction
The total force should be equal to the centripetal force by Newton's Second Law.

where
because we are looking for the case where the car loses contact.

Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.

Explanation:
The point that might confuse you in this question is the direction of the normal force at the top of the loop.
We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.
Each stream in a drainage system drains into a certain area. In a drainage basin the water falling in the basin drain will fall into the same stream. A drainage divides drawing basin from other drainage basins
Answer:
The initial speed of the pelican is 8.81 m/s.
Explanation:
Given;
height of the pelican, h = 5.0 m
horizontal distance, X = 8.9 m
The time of flight is given by;

The initial horizontal speed of the pelican is given by;
X = vₓt
vₓ = X / t
vₓ = 8.9 / 1.01
vₓ = 8.81 m/s
Therefore, the initial speed of the pelican is 8.81 m/s.