Sun warms the earth with Electromagnetic Radiations.
Sun emits out short range radiation in the form of Electromagnetic (EM) Radiation. Our Atmosphere is transparent to EM radiation. It does not heats up the atmosphere but when it collide with the solid objects or particles it is absorbed, reflected or refracted and the object heats up and start emitting radiation in the form of long wave radiation. Our Earth does the same. The long wave radiation emitted by the earth surface is absorbed by our atmosphere and it heats up. Therefore, we can say our atmosphere is heated from below.
<em>L</em><em>ike the alto in a choir the octavina plays the lower melody . what musical element does this tell?</em>
answer : melody....
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:
It only depends on the vertical component
Explanation:
Hello!
The horizontal component will tell you how much you travel in that direction.
You could have a large horizontal velocity, but if the vertical velocity is zero, you will never be out of the ground. Similarly, you could have a zero horizontal velocity, but if you have a non-zero vertical velocity you will be some time off the ground. This time can be calculated by two means, one is using the equation of motion (position as a function of time) and the other using the velocity as a fucntion of time.
For the former you must find the time when the position is zero.
Lets consider the origin of teh coordinate system at your feet
y(t) = vt - (1/2)gt^2
We are looking for a time t' for which y(t')=0
0 = vt' - (1/2)gt'^2
vt' = (1/2)gt'^2
The trivial solution is when t'=0 which is the initial position, however we are looking for t'≠0, therefore we can divide teh last equation by t'
v = (1/2)gt'
Solving for t'
t' = (2v/g)