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