Answer:
h = 3.1 cm
Explanation:
Given that,
The volume of a oil drop, V = 10 m
Radius, r = 10 m
We need to find the thickness of the film. The film is in the form of a cylinder whose volume is as follows :
So, the thickness of the film is equal to 3.1 cm.
Answer:
5.5 km
Explanation:
First, we convert the distance from km/h to m/s
910 * 1000/3600
= 252.78 m/s
Now, we use the formula v²/r = gtanθ to get our needed radius
making r the subject of the formula, we have
r = v²/gtanθ, where
r = radius of curvature needed
g = acceleration due to gravity
θ = angle of banking
r = 252.78² / (9.8 * tan 50)
r = 63897.73 / (9.8 * 1.19)
r = 63897.73 / 11.662
r = 5479 m or 5.5 km
Thus, we conclude that the minimum curvature radius needed for the turn is 5.5 km
Given:
Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min
Required:
Inlet flowrate
Solution:
The problem can be solved by this general formula.
Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
First, we need to convert the units of the accumulation velocity into m/s to be consistent.
Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s
We then calculate the area of the pool and the area of the orifice by:
Area of pool = 3 × 4 m²
Area of pool = 12m²
Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²
Since we have all we need, we plug in the values to the general equation earlier
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²
Transposing terms,
Inlet flowrate = 0.316 m³/s
Wave speed = (wavelength) x (frequency)
= (4 m) x (2 /sec)
= 8 m/sec
Answer:
"0.758315 sec" is the appropriate choice.
Explanation:
The given values are:
Angular speed,
= 29 rad/s
or,
= 
Tire rotates,
= 3.5 times
Now,
The time interval will be:
⇒ 
⇒ 
⇒ 
⇒ 
