I believe if the objects speed increases!
Answer:
The time required for sucrose transportation through the tube is 8.4319 sec.
Explanation:
Given:
L = 0.025 m
A = 6.5×10^-4 m^2
D = 5×10^-10 m^2/s
ΔC = 5.2 x 10^-3 kg/m^3
m = 5.7×10^-13 kg
Solution:
t = m×L / D×A×ΔC
t = (5.7×10^-13) × (0.025) / (5×10^-10)×(6.5×10^-4)×(5.2 x 10^-3)
t = 8.4319 sec.
Answer:
At least 6 N, assuming that the rope has a length of more than 2 meters.
Explanation:
In general, if it takes a force of newtons to lift an object at height , the work done lifting the object from to can be found using the definite integral about :
.
If the value of is a constant regardless of height , then the result of the integral would be
.
However, in this case the value of does depend on the value of .
- At height , nothing is being lifted. The amount of force required would be zero.
- At height , one meter of the rope is in the air. That requires a force of at least .
- In general, at a height of meters, the force required would be at least Newtons.
In other words, where is in Newtons and is in meters.
Evaluate the integral:
.
Most reactions are exothemic. If the forward reaction of an equilibrium reaction is exothemic then the reverse reaction must be endothermic.
If a system in equilibrium is heated, it will move in exothermic direction to give out heat energy.
<u>Answer:</u>
Given Data:
V2 ?
V1 = 200 ml,
P1 = 500kpa and P2 = 2500kpa
From <em>Boyels law</em> (Ideal gas law) Where temperature remains constant, <em>Pressure is inversely proportional to volume.</em>
P1 .V1 = P2 . V2 ; T=Constant
V2 = (P1.V1) ÷ P2
= (500×200) ÷(250)
= 400 ml
<em>The final volume of the gas is 400 ml.</em>
<em>Note: Increased pressure decreases the volume or decreased pressure increase the volume as they are inversely proportional.</em><em> Here in our answer pressure decreases from 500 kpa to 250 kpa so volume increases from 200 ml to 400 ml. </em>