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
Answer:
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Answer:
52 mm/s (approximately)
Explanation:
Given:
Initial speed of the projectile is, 
Angle of projection is, 
Time taken to land on the hill is, 
In a projectile motion, there is acceleration only in the vertical direction which is equal to acceleration due to gravity acting vertically downward. There is no acceleration in the horizontal direction.
So, the velocity in the horizontal direction always remains the same.
The horizontal component of initial velocity is given as:
Now, the velocity in the vertical direction goes on decreasing and becomes 0 at the highest point of the trajectory. So, at the highest point, only horizontal component acts.
Therefore, the projectile's velocity at the highest point of its trajectory is equal to the horizontal component of initial velocity and thus is equal to 52 mm/s.
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The answer would be increased number <span />of rod cells.Hope this helps :)
