Answer:
1.12 m
0.08291 m
Explanation:
u = Upstream velocity = 0.4 m/s
Re = Reynold's number =
(turbulent)
= Viscosity of water = 
Here the flow is turbulent so we have the relation

The approximate location downstream from the leading edge where the boundary layer becomes turbulent is 1.4 m
Boundary layer thickness relation is given by

The boundary layer thickness is 0.08291 m
Answer:
m =
x,
graph of x vs m
Explanation:
For this exercise, the simplest way to determine the mass of the cylinder is to take a spring and hang the mass, measure how much the spring has stretched and calculate the mass, using the translational equilibrium equation
F_e -W = 0
k x = m g
m =
x
We are assuming that you know the constant k of the spring, if it is not known you must carry out a previous step, calibrate the spring, for this a series of known masses are taken and hung by measuring the elongation (x) from the equilibrium position, with these data a graph of x vs m is made to serve as a spring calibration.
In the latter case, the elongation measured with the cylinder is found on the graph and the corresponding ordinate is the mass
Answer:
C. The warm air will rise, causing water vapor to condense and form clouds.
Explanation:
Answer:
Part a)

Part b)
this is a large potential which can not be possible because at this high potential the air will break down and the charge on the sphere will decrease.
Part C)
here we can assume the sphere is placed at vacuum so that there is no break down of air.
Explanation:
Part a)
As we know that the potential near the surface of metal sphere is given by the equation

here we have
Q = 8 C
R = 10.0 cm
now we have


Part b)
this is a large potential which can not be possible because at this high potential the air will break down and the charge on the sphere will decrease.
Part C)
here we can assume the sphere is placed at vacuum so that there is no break down of air.
I believe it'd accelerate at 1.25 m/s^2 instead of 1, as it lost 1/4 of its mass (.25), so now it is .25 of 1 faster.