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.
In electricity, the most famous and basic equation is the Ohm's Law which relates the parameters voltage, current and resistance. One form of this law as written in equation is V = IR, where V is the voltage in volts, I is the current in amperes and R is the resistance in ohms. These parameters depends in the arrangements, whether it's series or parallel.
In a series connection, the voltage is greater across a high-resistance resistor. Therefore, the voltage is much greater for the 20-ohm resistor. However,if it is a parallel circuit, the voltage is just the same for both resistors.
(1) The harmonic number for the mode of oscillation is 3.
(2) The pitch (frequency) of the sound is 579.55 Hz
(3) The level of the water inside the vertical pipe is 0.1 m.
<h3>The harmonic number</h3>
The harmonic number for the mode of oscillation illustrated for the closed pipe is 3.
<h3>Frequency of the wave</h3>
The pitch (frequency) of the sound is calculated from third harmonic formula;
f = 3v/4L
where;
- v is speed of sound
- L is length of the pipe
f = (3 x 340) / (4 x 0.44)
f = 579.55 Hz
<h3>level of the water</h3>
wave equation for first harmonic of a closed pipe is given as
f = v/(4L)
251.1 = 340/(4L)
4L = 340/251.1
4L = 1.35
L = 1.35/4
L = 0.34 m
level of water = 0.44 m - 0.34 m = 0.1 m
Thus, the level of the water inside the vertical pipe is 0.1 m.
Learn more about harmonics of closed pipes here: brainly.com/question/27248821
#SPJ1
Answer:
The heat loss per unit length is 
Explanation:
From the question we are told that
The outer diameter of the pipe is 
The thickness is 
The temperature of water is 
The outside air temperature is 
The water side heat transfer coefficient is 
The heat transfer coefficient is 
The heat lost per unit length is mathematically represented as
![\frac{Q}{L} = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1} + \frac{ln [\frac{d}{D} ]}{z_2}}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7BL%7D%20%20%20%3D%20%5Cfrac%7B2%20%5Cpi%20%28T%20-%20Ta%29%7D%7B%20%5Cfrac%7Bln%20%5B%5Cfrac%7Bd%7D%7BD%7D%20%5D%7D%7Bz_1%7D%20%20%2B%20%20%5Cfrac%7Bln%20%5B%5Cfrac%7Bd%7D%7BD%7D%20%5D%7D%7Bz_2%7D%7D)
Substituting values
![\frac{Q}{L} = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300} + \frac{ln [\frac{0.104}{0.002} ]}{20}}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7BL%7D%20%20%20%3D%20%5Cfrac%7B2%20%2A%203.142%20%28363%20-%20263%29%7D%7B%20%5Cfrac%7Bln%20%5B%5Cfrac%7B0.104%7D%7B0.002%7D%20%5D%7D%7B300%7D%20%20%2B%20%20%5Cfrac%7Bln%20%5B%5Cfrac%7B0.104%7D%7B0.002%7D%20%5D%7D%7B20%7D%7D)
