Normally, the water pressure inside a pump is higher than the vapor pressure: in this case, at the interface between the liquid and the vapor, molecules from the liquid escapes into vapour form. Instead, when the pressure of the water becomes lower than the vapour pressure, molecules of vapour can go inside the water forming bubbles: this phenomenon is called
cavitation.
So, cavitation occurs when the pressure of the water becomes lower than the vapour pressure. In our problem, vapour pressure at
is 1.706 kPa. Therefore, the lowest pressure that can exist in the pump without cavitation, at this temperature, is exactly this value: 1.706 kPa.
Choice C.
That's when convection stops.
Answer:
The 10X objective is use for the identification of actual size of histology tissues and 4X magnification is best for observation of most tissues slides
Explanation:
4X magnification is best for observation of most tissues slides because it has an objective lens that have lower power and have great high field overview which make it very easier to locate specimens on the slide. It is use to get the overview of histology slides. It is use to showcase more detailed observations about histology.
The 40X objective is use majorly to identify tissue , to observe the finer details and study tissue organization on the histology slide.
Answer:
3.28 m
3.28 s
Explanation:
We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.
Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
X0 = 0
V0 = 4 m/s
a = -2.45 m/s^2 (because the acceleration is down slope)
Then:
X(t) = 4*t - 1.22*t^2
And the equation for speed is:
V(t) = V0 + a * t
V(t) = 4 - 2.45 * t
If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:
0 = 4 - 2.45 * t
4 = 2.45 * t
t = 1.63 s
Replacing that time on the position equation:
X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m
To find the time it will take to return we equate the position equation to zero:
0 = 4 * t - 1.22 * t^2
Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:
0 = t * (4 - 1.22*t)
t1 = 0
0 = 4 - 1.22*t2
1.22 * t2 = 4
t2 = 3.28 s
