Define
v = volume of a drop per second, cm³/s
The time taken to fill 200 cm³ is 1 hour.
Let V = 200 cm³, the filled volume.
Let t = 1 h = 3600 s, the time required to fill the volume.
Therefore,

The average volume of a single drop is approximately 0.0556 cm³.
Answer: 0.0556 cm³
Answer:
25.71 kgm/s
Explanation:
Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.
Given that K₂ = 0.7K₁
1/2mv₂² = 0.7(1/2mv₁²)
v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s
Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.
Th magnitude of object A's momentum change is thus 25.71 kgm/s
Answer:
m = 2.01[kg]
Explanation:
This problem can be solved using Newton's second law which tells us that the force applied on a body is equal to the product of mass by acceleration.

where:
F = force = 12.5 [N]
m = mass [kg]
a = acceleration = 6.2 [m/s²]
![12.5=m*6.2\\m = 2.01[kg]](https://tex.z-dn.net/?f=12.5%3Dm%2A6.2%5C%5Cm%20%3D%202.01%5Bkg%5D)
Answer:
m = 56.5 kg
Explanation:
Since the addition of mass on one piston caused a change in pressure head at the other. Diameter of the piston calculated is used as 0.46 m
Δm*g / Area = p * g * Δh ..... Eq1

