Answer:
As you know, the denser objects have more weight per unit of volume, this will mean that the force that pulls down these objects is a bit larger.
This will mean that the denser objects will always go to the bottom.
This clearly implies that the red liquid, the one with one of the smaller densities, can not be at the bottom.
There are some cases where a liquid with a small density may become a lot denser as the temperature or pressure changes, and in a case like that, we could see the red liquid at the bottom, but for this case, there is no mention of changes in the temperature nor in the pressure, so this can be discarded.
The only thing that makes sense is that the red part at the bottom is the base of the tube, and has nothing to do with the red liquid.
Answer:
The answer is
<h2>84.9 kPa</h2>
Explanation:
Using Boyle's law to find the final pressure
That's
where
P1 is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume
Since we are finding the final pressure
From the question
P1 = 115 kPa
V1 = 480 mL
V2 = 650 ml
So we have
We have the final answer as
<h3>84.9 kPa</h3>
Hope this helps you
Refer to the diagram shown below.
The hoist is in static equilibrium supported by tensions in the two ropes.
For horizontal force balance, obtain
T₃ cos 50 = T₂ cos 38
0.6428T₃ = 0.788T₂
T₃ = 1.2259T₂ (1)
For vertical force balance, obtain
T₂ sin 38 + T₃ sin 50 = 350
0.6157T₂ + 0.766T₃ = 350 (2)
Substitute (1) into (2).
0.6157T₂ + 0.766(1.2259T₂) = 350
1.5547T₂ = 350
T₂ = 225.124 N
T₃ = 1.2259(225.124) = 275.979
Answer:
T₂ = 225.12 N
T₃ = 275.98 N
Answer:
0.159
Explanation:
the formula to find its is 1÷2*gt^2