Explanation:
Given:
m = 1.673 × 10^-27 kg
Q = q = 1.602 × 10^-19 C
r = 0.75 nm
= 0.75 × 10^-9 m
A.
Energy, U = (kQq)/r
Ut = 1/2 mv^2 + 1/2 mv^2
1.673 × 10^-27 × v^2 = (8.99 × 10^9 × (1.602 × 10^-19)^2)/0.75 × 10^-9
v = 1.356 × 10^4 m/s
B.
F = (kQq)/r^2
F = m × a
1.673 × 10^-27 × a = ((8.99 × 10^9 × (1.602 × 10-19)^2)/(0.075 × 10^-9)^2
a = 2.45 × 10^17 m/s^2.
Answer:
a) The mass flow rate through the nozzle is 0.27 kg/s.
b) The exit area of the nozzle is 23.6 cm².
Explanation:
a) The mass flow rate through the nozzle can be calculated with the following equation:
Where:
: is the initial velocity = 20 m/s
: is the inlet area of the nozzle = 60 cm²
: is the density of entrance = 2.21 kg/m³
Hence, the mass flow rate through the nozzle is 0.27 kg/s.
b) The exit area of the nozzle can be found with the Continuity equation:
Therefore, the exit area of the nozzle is 23.6 cm².
I hope it helps you!
45mph is the answer if you do the math right
Answer:
Earth's atmosphere traps energy from the sun. Which is a direct result of the trapping of energy by earth's atmosphere? Earth has moderate temperatures.
Explanation:
