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:
7503.13 N/m
Explanation:
Use principle of conservation of energy.
Here, energy stored in the spring due to compression shall be utilized in attaining the potential energy of the mug.
Given that,
Length of the spring = 20 cm = 0.20 m
Compression, x = 8 cm = 0.08 m
mass of the mug, m = 350 g = 0.35 kg
h = 7 m
use the expression for energy balance -
(1/2)*k*x^2 = m*g*h
=> k = (2*m*g*h) / x^2
input the values
k = (2*0.35*9.8*7) / 0.08^2
= 7503.13 N/m
Mass number is the total of neutrons and protons added together therefore it is 34+40 which is 74. This is the mass number for the atom.
