The volume flow rates for ∆P is 6.81m³/s .
<h3>What is pressure?</h3>
The amount of force applied on perpendicular to the surface of an object per unit area. The unit of it is pascal.
According to bernaulli's theorem theorem
P+1/2pV²+pgy = constant
where p fluid density
g is acceleration due to gravity, pressure at elevation,v is Velocity at elevation ,y is height of elevation.
As there are two tubes then the height of tube 1 is equal to height of tube two .
P1-P2=1/2p(Vd²-Vl²)
The flow rate of liquid is A1V1=A2V2 .
rest is attached in image
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Answer: I didn't see a difference because the large ball's vertical displacement and velocity are the same as the small one's.
Explanation:
A small 20-kg canoe is floating downriver at a speed of 2 m/s. 40 J is the canoe’s kinetic energy.
Answer: Option A
<u>Explanation:</u>
The given canoe has the mass and is being given to move at a speed. Therefore the kinetic energy of the canoe can be calculated using the following method,
Given that mass of the canoe = 20 kg and its speed =1 m/s
As we know that the Kinetic energy has the formula,

Therefore, substituting the value into the equation, we get,
= 40 J
Answer:
Incomplete question: "Each block has a mass of 0.2 kg"
The speed of the two-block system's center of mass just before the blocks collide is 2.9489 m/s
Explanation:
Given data:
θ = angle of the surface = 37°
m = mass of each block = 0.2 kg
v = speed = 0.35 m/s
t = time to collision = 0.5 s
Question: What is the speed of the two-block system's center of mass just before the blocks collide, vf = ?
Change in momentum:




It is neccesary calculate the force:

Here, g = gravity = 9.8 m/s²


Answer: 1477.78 N
Explanation:
Let's assume that the cross sectional area of the smaller piston be A1
let's also assume the cross sectional area of the larger piston be A2
We assume the force applied to the smaller piston be F1
We also assume the force applied to the larger piston be F2
we then use the formula
F1/A1 = F2/A2
From our question,
The radius of the smaller piston is 5 cm = 0.05 m
The radius of the larger piston is 15 cm = 0.15 m
The force of the larger piston is 13300 N
The force of the smaller piston is unknown = F
A1 = πr² = 3.142 * 0.05² = 0.007855 m²
A2 = πr² = 3.142 * 0.15² = 0.070695 m²
F1/0.007855 = 13300/0.070695
F1 = (13300 * 0.007855) / 0.070695
F1 = 104.4715 / 0.070695
F1 = 1477.78 N
Thus, the force the compressed air must exert is 1477.78 N