<h2><u>Answer:</u></h2>
As you are looking for a new tennis partner. People should keep in mind that they should go for the one who most likely demonstrate good sportsmanship
Luis, when you pursue the principles in tennis, you realize when to talk up, you don't blast a racquet or shout, holler.
Whatever it is following the principles and being respectful it the most ideal approach.
Hello there.
Question: <span>Which is an example of a covalent bond?
Answer: Water and Diamonds are good examples of covalent bonds.
Hope This Helps You!
Good Luck Studying ^-^</span>
Answer:
a
b

Explanation:
From the question we are told that
The pressure of the manometer when there is no gas flow is 
The level of mercury is 
The drop in the mercury level at the visible arm is 
Generally when there is no gas flow the pressure of the manometer is equal to the gauge pressure which is mathematically represented as

Here
is the density of mercury with value 
and
is the difference in the level of gas in arm one and two
So


Generally the height of the mercury at the arm connected to the pipe is mathematically represented as

=> 
Generally from manometry principle we have that
![P_G + \rho * g * d - \rho * g * [h - (h_m + d)] = 0](https://tex.z-dn.net/?f=P_G%20%2B%20%5Crho%20%2A%20g%20%20%2A%20d%20%20%20-%20%20%5Crho%20%2A%20%20g%20%20%2A%20%5Bh%20-%20%28h_m%20%2B%20d%29%5D%20%3D%200)
Here
is the pressure of the gas
![P_G +13.6 *10^{3} * 9.8 * 0.039 - 13.6 *10^{3} * 9.8 * [0.950 - (0.148 + 0.039)] = 0](https://tex.z-dn.net/?f=P_G%20%2B13.6%20%2A10%5E%7B3%7D%20%2A%209.8%20%20%2A%200.039%20%20%20%20-%20%2013.6%20%2A10%5E%7B3%7D%20%20%2A%20%209.8%20%20%2A%20%5B0.950%20-%20%280.148%20%2B%200.039%29%5D%20%3D%200)

converting to psig
One thing thats the same is there both universes.
One thing diffrent is that in the milky way there is proven life on earth but ursa major is proven without life.
Answer:
Explanation:
This question is based on the Law of Conservation of Angular Momentum.
Angular momentum (L) equals the moment of inertia (I) times the angular speed (ω).
L = Iω
If momentum is conserved,
I₁ω₁ = I₂ω₂
Data:
I₁ = 3.5 kg·m²s⁻¹
ω₁ = 6.0 rev·s⁻¹
I₂ = 0.70 kg·m²s⁻¹
Calculation:
