Explanation:
Given that,
Frequency of train horn, f = 218 Hz
Speed of train, ![v_t = 31.7 m/s](https://tex.z-dn.net/?f=v_t%20%3D%2031.7%20m%2Fs)
The speed of sound, V = 344 m/s (say)
The speed of the observed person, ![V_o=0\ m/s](https://tex.z-dn.net/?f=V_o%3D0%5C%20m%2Fs)
(a) When the train approaches you, the Doppler's effect gives the frequency as follows :
![f'=f(\dfrac{V}{V-v_t})\\\\f'=218\times (\dfrac{344}{344-31.7})\\\\f'=240.12\ Hz](https://tex.z-dn.net/?f=f%27%3Df%28%5Cdfrac%7BV%7D%7BV-v_t%7D%29%5C%5C%5C%5Cf%27%3D218%5Ctimes%20%28%5Cdfrac%7B344%7D%7B344-31.7%7D%29%5C%5C%5C%5Cf%27%3D240.12%5C%20Hz)
(b) When the train moves away from you, the Doppler's effect gives the frequency as follows :
![f'=f(\dfrac{V}{V+v_t})\\\\f'=218\times (\dfrac{344}{344+31.7})\\\\f'=199.6\ Hz](https://tex.z-dn.net/?f=f%27%3Df%28%5Cdfrac%7BV%7D%7BV%2Bv_t%7D%29%5C%5C%5C%5Cf%27%3D218%5Ctimes%20%28%5Cdfrac%7B344%7D%7B344%2B31.7%7D%29%5C%5C%5C%5Cf%27%3D199.6%5C%20Hz)
Hence, this is the required solution.
(a) Force between the two charges
The electrostatic force between the two charges is given by:
![F=k\frac{q_1 q_2}{r^2}](https://tex.z-dn.net/?f=%20F%3Dk%5Cfrac%7Bq_1%20q_2%7D%7Br%5E2%7D%20%20)
where k is the Coulomb's constant, q1 and q2 the two charges, r their separation.
In this problem:
![q_1 =6 \mu C=6 \cdot 10^{-6}C](https://tex.z-dn.net/?f=%20q_1%20%3D6%20%5Cmu%20C%3D6%20%5Ccdot%2010%5E%7B-6%7DC%20)
![q_2=2 \mu C=2 \cdot 10^{-6}C](https://tex.z-dn.net/?f=%20q_2%3D2%20%5Cmu%20C%3D2%20%5Ccdot%2010%5E%7B-6%7DC%20)
![r=0.1 m](https://tex.z-dn.net/?f=%20r%3D0.1%20m%20)
Substituting into the equation, we find
![F=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(6 \cdot 10^{-6}C)(2 \cdot 10^{-6}C)}{(0.1 m)^2}=10.8 N](https://tex.z-dn.net/?f=%20F%3D%288.99%20%5Ccdot%2010%5E9%20Nm%5E2C%5E%7B-2%7D%29%5Cfrac%7B%286%20%5Ccdot%2010%5E%7B-6%7DC%29%282%20%5Ccdot%2010%5E%7B-6%7DC%29%7D%7B%280.1%20m%29%5E2%7D%3D10.8%20N%20%20)
(b) direction of particle q2
Particle q2 wants to move in the direction of the force acting on it. The direction of the force depends on the relative sign of the two charges: like charges attract each other, opposite charges repel each other. In this case, the two charges are both positive, so they repel each other and q2 tends to move away from particle q1.
Whatever the statement is, I know one thing for sure: It's NOT included on the list of choices that you've provided.
Answer:
h = 9.83 cm
Explanation:
Let's analyze this interesting exercise a bit, let's start by comparing the density of the ball with that of water
let's reduce the magnitudes to the SI system
r = 10 cm = 0.10 m
m = 10 g = 0.010 kg
A = 100 cm² = 0.01 m²
the definition of density is
ρ = m / V
the volume of a sphere
V =
V =
π 0.1³
V = 4.189 10⁻³ m³
let's calculate the density of the ball
ρ =
ρ = 2.387 kg / m³
the tabulated density of water is
ρ_water = 997 kg / m³
we can see that the density of the body is less than the density of water. Consequently the body floats in the water, therefore the water level that rises corresponds to the submerged part of the body. Let's write the equilibrium equation
B - W = 0
B = W
where B is the thrust that is given by Archimedes' principle
ρ_liquid g V_submerged = m g
V_submerged = m / ρ_liquid
we calculate
V _submerged = 0.10 9.8 / 997
V_submerged = 9.83 10⁻⁴ m³
The volume increassed of the water container
V = A h
h = V / A
let's calculate
h = 9.83 10⁻⁴ / 0.01
h = 0.0983 m
this is equal to h = 9.83 cm
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:
![\dot{m_{i}} = \rho_{i} v_{i}A_{i}](https://tex.z-dn.net/?f=%20%5Cdot%7Bm_%7Bi%7D%7D%20%3D%20%5Crho_%7Bi%7D%20v_%7Bi%7DA_%7Bi%7D%20)
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:
![\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}](https://tex.z-dn.net/?f=%20%5Crho_%7Bi%7D%20v_%7Bi%7DA_%7Bi%7D%20%3D%20%5Crho_%7Bf%7D%20v_%7Bf%7DA_%7Bf%7D%20)
![0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}](https://tex.z-dn.net/?f=%200.27%20kg%2Fs%20%3D%200.762%20kg%2Fm%5E%7B3%7D%2A150%20m%2Fs%2AA_%7Bf%7D%20)
![A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}](https://tex.z-dn.net/?f=%20A_%7Bf%7D%20%3D%20%5Cfrac%7B0.27%20kg%2Fs%7D%7B0.762%20kg%2Fm%5E%7B3%7D%2A150%20m%2Fs%7D%20%3D%200.00236%20m%5E%7B2%7D%2A%5Cfrac%7B%28100%20cm%29%5E%7B2%7D%7D%7B1%20m%5E%7B2%7D%7D%20%3D%2023.6%20cm%5E%7B2%7D%20)
Therefore, the exit area of the nozzle is 23.6 cm².
I hope it helps you!