Answer:
<u>0.04 °C⁻¹</u>
Explanation:
First, we need to calculate linear expansivity, then after finding that value, we can move on to finding the area expansivity.
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Finding Linear Expansivity :
⇒ α = Final length - Original length / (Original length × ΔT)
⇒ α = 9 - 4 / (4 × 70 - 20)
⇒ α = 5 / 5 × 50
⇒ α = <u>0.02</u>
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Finding Area Expansivity :
⇒ Area Expansivity = 2 × Linear Expansivity
⇒ β = 2 × α
⇒ β = 2 × 0.02
⇒ β = <u>0.04 °C⁻¹</u>
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
The farther apart the two objects, the weaker the gravitational attraction between them.
Answer:
Action: Gravity pulls on the ball.
Reaction: The ball falls to the ground.
Explanation:
