Answer:
v2 = 65 m/s
the speed of the water leaving the nozzle is 65 m/s
Explanation:
Given;
Water flows at 0.65 m/s through a 3.0 cm diameter hose that terminates in a 0.3 cm diameter nozzle
Initial speed v1 = 0.65 m/s
diameter d1 = 3.0 cm
diameter (nozzle) d2 = 0.3 cm
The volumetric flow rates in both the hose and the nozzle are the same.
V1 = V2 ........1
Volumetric flow rate V = cross sectional area × speed of flow
V = Av
Area = (πd^2)/4
V = v(πd^2)/4 ....2
Substituting equation 2 to 1;
v1(πd1^2)/4 = v2(πd2^2)/4
v1d1^2 = v2d2^2
v2 = (v1d1^2)/d2^2
Substituting the given values;
v2 = (0.65 × 3^2)/0.3^2
v2 = 65 m/s
the speed of the water leaving the nozzle is 65 m/s
Answer:
reviewing the opinion of the two students we see that neither is right, since when the kinetic energy increases the potential energy decreases by the same value
Explanation:
For this exercise we must use the law of conservation of energy.
Starting point. Resting electron
Em₀ = U = eV
the potential difference and the electric field are related
V = - E d
Final point. When leaving the electric field
= K = ½ m v²
Em₀ = Em_{f}
e E d = ½ m v²
From this expression we see that when an electron moves from the initial point to the final point, the potential energy must decrease, for the total energy to be constant.
When reviewing the opinion of the two students we see that neither is right, since when the kinetic energy increases the potential energy decreases by the same value
Answer:
its unbalanced and balanced
Explanation:
<span>On the surface of the Earth, the escape velocity is about 11.2 km/s, which is approximately 33 times the speed of sound (Mach 33) and several times the muzzle velocity of a rifle bullet (up to 1.7 km/s). However, at 9,000 km altitude in "space", it is slightly less than 7.1 km/s.</span>
It is in its ground state when its orbital electron is at its lowest energy amount.