Answer:
The flux (volume of water per unit time) through the hoop will also double.
Explanation:
The flux = volume of water per unit time = flow rate of water through the hoop.
The Flow rate of water through the hoop is proportional to the area of the hoop, and the velocity of the water through the hoop.
This means that
Flow rate = AV
where A is the area of the hoop
V is the velocity of the water through the hoop
This flow rate = volume of water per unit time = Δv/Δt =Q
From all the above statements, we can say
Q = AV
From the equation, if we double the area, and the velocity of the stream of water through the hoop does not change, then, the volume of water per unit time will also double or we can say increases by a factor of 2
Answer:
False
Explanation:
When the horizontal shear forces act on the surface there is transverse shear stress at a particular point which is equal in magnitude. Pure bending is less common than a non uniform bending because the beam is not in equilibrium.
Answer:
(N-1) × (L/2R) = (N-1)/2
Explanation:
let L is length of packet
R is rate
N is number of packets
then
first packet arrived with 0 delay
Second packet arrived at = L/R
Third packet arrived at = 2L/R
Nth packet arrived at = (n-1)L/R
Total queuing delay = L/R + 2L/R + ... + (n - 1)L/R = L(n - 1)/2R
Now
L / R = (1000) / (10^6 ) s = 1 ms
L/2R = 0.5 ms
average queuing delay for N packets = (N-1) * (L/2R) = (N-1)/2
the average queuing delay of a packet = 0 ( put N=1)
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Answer:
Explanation:
We can assume that the general formula for the drag force is given by:
And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:
Where w is the width and h the height.
The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:
If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:
We can assume the same drag coeeficient 
and we got:
1.7 ft =0.518 m
60 mph = 26.822 m/s
In order to find the drag coeffcient we ned to estimate the Reynolds number first like this:
And the value for the kinematic vicosity was obtained from the table of physical properties of the air under standard conditions.
Now we can find the aspect ratio like this:
And we can estimate the calue of 
from a figure.
And we can calculate the power difference like this:
