Answer:
v=20m/S
p=-37.5kPa
Explanation:
Hello! This exercise should be resolved in the next two steps
1. Using the continuity equation that indicates that the flow entering the nozzle must be the same as the output, remember that the flow equation consists in multiplying the area by the speed
Q=VA
for he exitt
Q=flow=5m^3/s
A=area=0.25m^2
V=Speed
solving for V
velocity at the exit=20m/s
for entry
2.
To find the pressure we use the Bernoulli equation that states that the flow energy is conserved.
where
P=presure
α=9.810KN/m^3 specific weight for water
V=speed
g=gravity
solving for P1
the pressure at exit is -37.5kPa
Answer:
a. I = 0.76 A
b. Z = 150.74
c. RL₁ = 34.41 , RL₂ = 602.58
d. RL₂ = 602.58
Explanation:
V₁ = 116 V , R₁ = 77.0 Ω , Vc = 364 V , Rc = 473 Ω
a.
Using law of Ohm
V = I * R
I = Vc / Rc = 364 V / 473 Ω
I = 0.76 A
b.
The impedance of the circuit in this case the resistance, capacitance and inductor
V = I * Z
Z = V / I
Z = 116 v / 0.76 A
Z = 150.74
c.
The reactance of the inductor can be find using
Z² = R² + (RL² - Rc²)
Solve to RL'
RL = Rc (+ / -) √ ( Z² - R²)
RL = 473 (+ / -) √ 150.74² 77.0²
RL = 473 (+ / -) (129.58)
RL₁ = 34.41 , RL₂ = 602.58
d.
The higher value have the less angular frequency
RL₂ = 602.58
ω = 1 / √L*C
ω = 1 / √ 602.58 * 473
f = 285.02 Hz
Answer:
2.48 m/s
Explanation:
We can use the kinematic equation,
s = ut +½at²
Where
s = displacement
u = initial velocity
t = time taken
a = acceleration
Using the equation in vertical direction,
321 = 0×t +½×g×t², u = 0 because initial vertical velocity is 0
We get t = 8.01 s
Using the equation in the horizontal direction,
52 = u×8.01 +½×0×(8.01)²,. a = 0 because no unbalanced force act on object in that direction
So u = 2.48 m/s
The answer is C,<span> The sum of all forces acting on the object is zero. hope that helps!!</span>
Explanation:
The acceleration of the scooter is 2.5 m/s
