Answer:
V = 0.30787 m³/s
m = 2.6963 kg/s
v2 = 0.3705 m³/s
v2 = 6.017 m/s
Explanation:
given data
diameter = 28 cm
steadily =200 kPa
temperature = 20°C
velocity = 5 m/s
solution
we know mass flow rate is
m = ρ A v
floe rate V = Av
m = ρ V
flow rate = V = 
V = Av = 
V = 
V = 0.30787 m³/s
and
mass flow rate of the refrigerant is
m = ρ A v
m = ρ V
m = 
= 
m = 2.6963 kg/s
and
velocity and volume flow rate at exit
velocity = mass × v
v2 = 2.6963 × 0.13741 = 0.3705 m³/s
and
v2 = A2×v2
v2 = 
v2 = 
v2 = 6.017 m/s
Remote?? maybe I’m not really sure
Answer:
Any engineering job would be good YOU should be the one choosing which job.
Explanation:
Engineering is a great outlet for the imagination, and the perfect field for independent thinkers.
Answer:
Explanation:
<u>Projectile Motion</u>
In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:
Where 
is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.
The horizontal and vertical distances are, respectively:
The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find
Using this time in the horizontal distance, we find the Range or maximum horizontal distance:
Let's solve for
This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:
Or equivalently:
Given Vo=37 m/s and R=70 m
And
