Hypothesis , I believe the answer is hypothesis
Answer:
The speed of water flow inside the pipe at point - 2 = 34.67 m / sec
Explanation:
Given data
Diameter at point - 1 = 3.2 cm
Velocity at point - 1 = 1.1 m / sec = 110 cm / sec
Diameter at point - 2 = 0.57 cm
Velocity at point - 2 = ??
We know that from the continuity equation the rate of flow is constant inside a pipe between two points.
Thus
⇒ 
× 
= 
× 
⇒ 
× 
× =
⇒ × = 
×
⇒ 
× 110 = 
×
⇒ = 3467 cm / sec
⇒ = 34.67 m / sec
Thus the speed of water flow inside the pipe at point - 2 = 34.67 m / sec
I'm not 100% sure, but I believe what you mean is when they eject the old propulsion motors. Yes, they land in the ocean and the US Navy retrieves them for later use.
Hi,
The force that acts on hydraulic machine is heavy therefore the content must be something that cannot be compressed by that kind of force, the gas can easily be compressed while a liquid is nearly impossible to.
Answer:
a). V = 3.13*10⁶ m/s
b). T = 1.19*10^-7s
c). K.E = 2.04*10⁵
d). V = 1.02*10⁵V
Explanation:
q = +2e
M = 4.0u
r = 5.94cm = 0.0594m
B = 1.10T
1u = 1.67 * 10^-27kg
M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg
a). Centripetal force = magnetic force
Mv / r = qB
V = qBr / m
V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27
V = 2.09088 * 10^-20 / 6.68 * 10^-27
V = 3.13*10⁶ m/s
b). Period of revolution.
T = 2Πr / v
T = (2*π*0.0594) / 3.13*10⁶
T = 1.19*10⁻⁷s
c). kinetic energy = ½mv²
K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²
K.E = 3.27*10^-14J
1ev = 1.60*10^-19J
xeV = 3.27*10^-14J
X = 2.04*10⁵eV
K.E = 2.04*10⁵eV
d). K.E = qV
V = K / q
V = 2.04*10⁵ / (2eV).....2e-
V = 1.02*10⁵V
