Answer:
At centroid
Explanation:
In the given equilateral triangle ABC with side of 1 m. The three equal charges e,e,e are placed at the A,B and C.
And the fourth charge 2e is put at point O which is called centroid.
Now we can calculate the distance AD by applying pythagorean theorem as,
Put the values and get.
Now calculate AO as,
And the sides BO=CO=AO.
Now Force can be calculated as
And similarly,
Now we can calculate resultant of in upward direction. as,
Therefore the resultant force on centroid O.
Therefore the fourth charge 2e should be placed on centroid so that it experience zero force.
Answer:
A) P1=2 [bar] , W=-12 [kJ]
B) P1=0.8 [bar] , W=-7.3303 [kJ]
C) P1=0.6077 [bar] , W=-6.4091 [kJ]
Explanation:
First, from the problem we know the following information:
V1=0.1 m^3
V2=0.04 m^3
P2=2 bar =200 kPa
The relation PV^n=constant means PV^n is a constant through all the process, so we can derive the initial pressure as:
a) To the case a) the constant n is equal to 0, we can calculate the initial pressure substituting n=0 in the previous expression, so:
The expression to calculate the work is:
If n=0:
Then:
The work is:
b) To the case b) the constant n is equal to 1, we can calculate the initial pressure substituting n=1 in the initial expression, so:
If n=1 then:
To calculate the work:
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Substituting:
c) To the case c) the constant n is equal to 1.3, we can calculate the initial pressure substituting n=1.3 in the initial expression, so:
First:
The work:
Substituting:
W=-6.4091 kJ
This is just testing your ability to recall that kinetic energy is given by:
<span>k.e. = ½mv² </span>
<span>where m is the mass and v is the velocity of the particle. </span>
<span>The frequency of the light is redundant information. </span>
<span>Here, you are given m = 9.1 * 10^-31 kg and v = 7.00 * 10^5 m/s. </span>
<span>Just plug in the values: </span>
<span>k.e. = ½ * 9.1 * 10^-31 * (7.00 * 10^5)² </span>
<span>k.e. = 2.23 * 10^-19 J
so it will be d:2.2*10^-19 J</span>
Answer:
The horizontal velocity is constant at 16 m/s.
After 1 sec since v = a t then 9.8 m/s^2 * 1 sec = 9.8 m/s for the vertical velocity
V = (Vx^2 + Vy^2)^1/2 = (16^2 + 9.8^2)^1/2 = 18.8 m/s