Inelastic collision happens when two objects joined and move together after the collision
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:
If we have large numbers (b is positive) or small numbers (b is negative), then this way ... 1, and V2i = 100 L, n2i = 5 + 2 + 1 = 8 in vessel 2. ... a good working substance in the barometer.
In the above problem, we need to find mass of the second child, so that the Center of Mass remains at the origin( pivot).
CM= m1r1+m2r2/m1+m2
0= 20*-2+16*r2/20+16
r2= 40/16
r2= +2.5 m
Answer:
Explanation:
Work is equal to the product of force and distance.
The box is being pushed with an effort (force) of 45 Newtons and the distance is 3.5 meters.
Substitute the values into the formula.
Multiply.
- 1 Newton meter is equal to 1 Joule
- Our answer of 157.5 N*m equals 157.5 J
<u>157.5 Joules </u> of work are done on the crate.
