The elevation in reservoir at the rate of flow using is 03m/s is 114m.
The Reynolds range is the ratio of inertial forces to viscous forces. The Reynolds variety is a dimensionless variety used to categorize the fluids structures in which the impact of viscosity is crucial in controlling the velocities or the flow sample of a fluid.
The reason of the Reynolds number is to get a few experience of the relationship in fluid glide between inertial forces (this is those that maintain going by using Newton's first law – an item in motion stays in movement) and viscous forces, this is people who cause the fluid to come back to a forestall because of the viscosity of the fluid.
calculation,
Let L = 100 m pipe
L1 = 150 m pipe
H f = friction losses
Using Reynolds number, relative roughness, friction co- effiicients and friction losses
Substitute the value in equation
Z = 110= 0.48= 3.54
Z = 114m
Therefore water surface elevation at reservoir is 114 meter.
Learn more about rate of flow here:-brainly.com/question/21630019
#SPJ4
To determine the force that acts on the mass, just multiply the mass by the gravitational field. Using the given data,
F = (2.50 kg)(14 N/kg) = 35 N
Therefore, the force that acts on the mass is equal to 35 N.
Answer:
v₁ = 2.48m/s, v₂ = 0.02m/s
Explanation:
Momentum p must be conserved. p = mv
1) First person throwing the snow ball. The momentum before the throw:
p = (65kg + 0.045kg) * 2.5 m/s
The momentum after the throw:
p = 65kg * v₁ + 0.045kg * 30m/s
Solving for the velocity v₁ of person 1:
v₁ = ((65kg + 0.045kg) * 2.5 m/s - 0.045kg * 30m/s) / 65kg = 2.48m/s
2) Second person catching the ball. The momentum before the catch:
p = 0.045kg * 30m/s + 60kg * 0m/s
The momentum after the catch:
p = (60kg + 0.045kg) * v₂
Solving for velocity v₂ of person 2:
v₂ = 0.045kg * 30m/s / (60kg + 0.045kg) = 0.02 m/s
In #8, the distance and the (magnitude of displacement) are equal, because he crawled in a straight line.
Displacement = (straight-line distance from start-point to end-point) in the direction from start to end, regardless of what route was actually followed.
Displacement = 5m, in the negative direction.
In #9 . . . distance will be the same. Displacement is going to be the same magnitude, but in the positive direction.
This is so simple that it's hard to talk about.
In #8, "What was the bug's distance ?". "Distance was 5 meters.". "What was the bug's displacement ?", "Displacement was 5 meters backwards."
In #9, What was the bug's distance ?". "Distance was 5 meters.". "What was the bug's displacement ?", "Displacement was 5 meters forward."
Answer:
12.0 V
Explanation:
Data :
Potential difference due to a single charge (+Q), E = 3.0 V
The Electric potential for the system of charges is given as:
![E=\frac{1}{4\pi \epsilon_o}[\Sigma\frac{Q}{r}]](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_o%7D%5B%5CSigma%5Cfrac%7BQ%7D%7Br%7D%5D)
for single charge, E = 3.0 V = ![\frac{1}{4\pi \epsilon_o}[\frac{Q}{r}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_o%7D%5B%5Cfrac%7BQ%7D%7Br%7D%5D)
->eq(1)
And for 4 charges:
![E=\frac{1}{4\pi \epsilon_o}[4\frac{Q}{r}]](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_o%7D%5B4%5Cfrac%7BQ%7D%7Br%7D%5D)
-eq(2)
from eq(1) and (2) we have
E = 4 × 3.0 V = 12 V
