Answer:
The minimum coefficient of friction is 0.666
Explanation:
Suppose, In a classic carnival ride, patrons stand against the wall in a cylindrical shaped room. Once the room gets spinning fast enough, the floor drops from the bottom of the room! Friction between the walls of the room and the people on the ride make them the “stick” to the wall so they do not slide down. In one ride, the radius of the cylindrical room is R = 7.6 m and the room spins with a frequency of 20.9 revolutions per minute.
Given that,
The normal force does not exceed 1.5 times each persons weight
We need to calculate the minimum coefficient of friction
Using balance equation
Where, N = normal force
Put the value into the formula
Hence, The minimum coefficient of friction is 0.666
The equal velocity approach for duct size assumes that the air velocity in each duct segment is the same.
How fast is the air moving through a duct?
The most common unit of air velocity (distance traveled in a unit of time) is feet per minute (FPM). The amount of air passing past a location in the duct per period of time may be calculated by multiplying the airflow by the area of the duct. The standard unit for volume flow is cubic feet per minute (CFM).
What happens when the size of ducts changes to the airflow?
- Result for an image The equal velocity technique for duct size makes the assumption that air velocity is constant across the entire duct system.
- The main lesson to be learned from this is that when air goes from a bigger to a narrower duct, its velocity rises. The velocity drops when it transitions from a shorter to a bigger duct. The flow rate or the amount of air passing through the duct in cubic feet per minute is the same in all scenarios.
Learn more about air velocity here:
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Answer:
Approximate height of the building is 23213 meters.
Explanation:
Let the height of the building be represented by h.
0.02 radians = 0.02 × 
= 0.02 x (180/
)
0.02 radians = 1.146°
10.5 km = 10500 m
Applying the trigonometric function, we have;
Tan θ = 
So that,
Tan 1.146° = 
⇒ h = Tan 1.146° x 10500
= 2.21074 x 10500
= 23212.77
h = 23213 m
The approximate height of the building is 23213 m.
