Answer:
Flow Rate = 80 m^3 /hours (Rounded to the nearest whole number)
Explanation:
Given
- Hf = head loss
- f = friction factor
- L = Length of the pipe = 360 m
- V = Flow velocity, m/s
- D = Pipe diameter = 0.12 m
- g = Gravitational acceleration, m/s^2
- Re = Reynolds's Number
- rho = Density =998 kg/m^3
- μ = Viscosity = 0.001 kg/m-s
- Z = Elevation Difference = 60 m
Calculations
Moody friction loss in the pipe = Hf = (f*L*V^2)/(2*D*g)
The energy equation for this system will be,
Hp = Z + Hf
The other three equations to solve the above equations are:
Re = (rho*V*D)/ μ
Flow Rate, Q = V*(pi/4)*D^2
Power = 15000 W = rho*g*Q*Hp
1/f^0.5 = 2*log ((Re*f^0.5)/2.51)
We can iterate the 5 equations to find f and solve them to find the values of:
Re = 235000
f = 0.015
V = 1.97 m/s
And use them to find the flow rate,
Q = V*(pi/4)*D^2
Q = (1.97)*(pi/4)*(0.12)^2 = 0.022 m^3/s = 80 m^3 /hours
Answer:
The frequency of radiation is 
Explanation:
Given:
Wavelength 
m
Speed of light 
For finding the frequency of radiation,
Therefore, the frequency of radiation is
I believe it's B. the transmission of heat across matter
Answer
given,
mass of the block = 200 g = 0.2 Kg
Velocity at A = 0 m/s
Velocity at B = 8 m/s
slide to the horizontal distance = 10 m
height of the block be = 4 m
potential energy of the block
P = m g h
P = 0.2 x 9.8 x 4
P =7.84 J
kinetic energy
Work = P - KE
work = 7.84 - 6.14
work = 1.7 J
b) v² = u² + 2 a s
0 = 8² - 2 x a x 10
a = 3.2 m/s²
ma - μ mg = 0
Answer:
that technician A is right
Explanation:
The test lights are generally small bulbs that are turned on by the voltage and current flowing through the circuit in analog circuits, these two values are high and can light the bulb. In digital circuits the current is very small in the order of milliamps, so there is not enough power to turn on these lights.
From the above it is seen that technician A is right
