Use VFR1 = VFR2 to discover the velocity at in the hose VFR =
A * V
D hose =10 * D nozzle, R hose = 5 * D nozzle
Area of a circle = πR^2
Area h=3.14*25*D^2 = 75.5D^2
(Radius=Diameter/2) area n = 3.14*(D^2/4) = .785D^2
Use VFR = VFR v2 = 0.4m/s
0.4*.785D^2 = 75.5*D^2* v1 D^2
= .314 =75.5*V1
v1 = 0.004m/s
Now we have the velocity, we can use Bernoulli's equation.
P1+ρgh1+ρV1^2 /2 = constant
There is no atmospheric pressure before so the P1= the gauge
pressure at the pump, let’s call the height of the hose 0m and the height of
the nozzle 1m so the is no ρgh1 Likewise, there is only atmospheric pressure at
the nozzle which is 100000 PA, and lastly the density ρ of water is 1000 KG/M^3
Pg + 1000*.004^2/2 = 100000+1000*9.8*1+ 1000*0.4^2/2
Pg + .008= 100000+9800+80
Pg+.008= 109880
Pg=109880.008 PA
Answer: $85.80
Explanation: 100kg is equalvilent to 220 lbs, then you multiply 220 by .39 and then you have your answer :)
Answer:
12N
Explanation:
We are given that one of the forces are acting only in the horizontal x-direction. As a force must be applied on an object of mass in order to cause acceleration, the 6.0ms^-2 acceleration is due to the non-horizontal force acting on the 2.0kg object.
Using Newton's Second Law of motion; we know that for a constant mass, force is equal to mass times acceleration, F=ma.
Assuming the other force is acting only in the vertical direction (question doesn't specify, thus we are finding the minimum force to cause this acceleration):
F= 2.0kg * 6.0ms^-2
F=12.0 kgms^-2
F=12 N