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,
![AD^{2}=AB^{2}+BD^{2}](https://tex.z-dn.net/?f=AD%5E%7B2%7D%3DAB%5E%7B2%7D%2BBD%5E%7B2%7D)
Put the values and get.
![AD^{2}=1^{2}+(\frac{1}{2} )^{2}\\AD=\sqrt{\frac{3}{4} } \\AD=\frac{\sqrt{3}}{2}](https://tex.z-dn.net/?f=AD%5E%7B2%7D%3D1%5E%7B2%7D%2B%28%5Cfrac%7B1%7D%7B2%7D%20%29%5E%7B2%7D%5C%5CAD%3D%5Csqrt%7B%5Cfrac%7B3%7D%7B4%7D%20%7D%20%5C%5CAD%3D%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D)
Now calculate AO as,
![AO=\frac{2}{3}\times \frac{\sqrt{3} }{2}\\AO=\frac{1}{\sqrt{3} }](https://tex.z-dn.net/?f=AO%3D%5Cfrac%7B2%7D%7B3%7D%5Ctimes%20%5Cfrac%7B%5Csqrt%7B3%7D%20%7D%7B2%7D%5C%5CAO%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%20%7D)
And the sides BO=CO=AO.
Now Force can be calculated as
![F_{1}=\frac{2ke^{2} }{\frac{1}{\sqrt{3} } ^{2} }\\F_{1}=6ke^{2}](https://tex.z-dn.net/?f=F_%7B1%7D%3D%5Cfrac%7B2ke%5E%7B2%7D%20%7D%7B%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%20%7D%20%5E%7B2%7D%20%7D%5C%5CF_%7B1%7D%3D6ke%5E%7B2%7D)
And similarly,
![F_{2}=F_{3}=6ke^{2}](https://tex.z-dn.net/?f=F_%7B2%7D%3DF_%7B3%7D%3D6ke%5E%7B2%7D)
Now we can calculate resultant of
in upward direction. as,
![F_{net}=\sqrt{F_{2}^{2}+F_{3}^{2}+2F_{2}F_{3}cos120 } \\F_{net}=\sqrt{F_{2}^{2}+F_{2}^{2}+2F_{2}F_{2}(-\frac{1}{2})}\\F_{net}=6ke^{2}](https://tex.z-dn.net/?f=F_%7Bnet%7D%3D%5Csqrt%7BF_%7B2%7D%5E%7B2%7D%2BF_%7B3%7D%5E%7B2%7D%2B2F_%7B2%7DF_%7B3%7Dcos120%20%20%7D%20%5C%5CF_%7Bnet%7D%3D%5Csqrt%7BF_%7B2%7D%5E%7B2%7D%2BF_%7B2%7D%5E%7B2%7D%2B2F_%7B2%7DF_%7B2%7D%28-%5Cfrac%7B1%7D%7B2%7D%29%7D%5C%5CF_%7Bnet%7D%3D6ke%5E%7B2%7D)
Therefore the resultant force on centroid O.
![F=F_{1}-F_{net}\\F=6ke^{2}-6ke^{2}\\F=0](https://tex.z-dn.net/?f=F%3DF_%7B1%7D-F_%7Bnet%7D%5C%5CF%3D6ke%5E%7B2%7D-6ke%5E%7B2%7D%5C%5CF%3D0)
Therefore the fourth charge 2e should be placed on centroid so that it experience zero force.
Answer:
The ball is making a circular motion. The centripetal acceleration in a circular motion is
![a = \frac{v^2}{r}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7Bv%5E2%7D%7Br%7D%20)
~
Explanation:
Normally in physics questions like this, we use all the information in the question. Here, we left out the mass or the height of the object. I assume there is more sub-questions that is not shared here. However, the centripetal acceleration can be found by using the velocity and the radius only.
To solve this problem it is necessary to apply the concepts related to Normal Force, frictional force, kinematic equations of motion and Newton's second law.
From the kinematic equations of motion we know that the relationship of acceleration, velocity and distance is given by
![v_f^2=v_i^2+2ax](https://tex.z-dn.net/?f=v_f%5E2%3Dv_i%5E2%2B2ax)
Where,
Final velocity
Initial Velocity
a = Acceleration
x = Displacement
Acceleration can be expressed in terms of the drag coefficient by means of
Frictional Force
Force by Newton's second Law
Where,
m = mass
a= acceleration
Kinetic frictional coefficient
g = Gravity
Equating both equation we have that
![F_f = F](https://tex.z-dn.net/?f=F_f%20%3D%20F)
![\mu_k mg=ma](https://tex.z-dn.net/?f=%5Cmu_k%20mg%3Dma)
![a = \mu_k g](https://tex.z-dn.net/?f=a%20%3D%20%5Cmu_k%20g)
Therefore,
![v_f^2=v_i^2+2ax](https://tex.z-dn.net/?f=v_f%5E2%3Dv_i%5E2%2B2ax)
![0=v_i^2+2(\mu_k g)x](https://tex.z-dn.net/?f=0%3Dv_i%5E2%2B2%28%5Cmu_k%20g%29x)
Re-arrange to find x,
![x = \frac{v_i^2}{2(-\mu_k g)}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7Bv_i%5E2%7D%7B2%28-%5Cmu_k%20g%29%7D)
The distance traveled by the car depends on the coefficient of kinetic friction, acceleration due to gravity and initial velocity, therefore the three cars will stop at the same distance.
Answer:
It will take both pumps 3.08 hours to fill the tank working together.
Explanation:
Pump A can fill the tank in 5 hours. Assuming that the pump gives out a steady flow of water, in one hour, pump A can fill 1/5th of the tank. Similarly, pump B in an hour, fills up 1/8th of the tank.
We must add up these two values, in order to find how much of the tank the two pumps can fill up together in one hour.
1/5 +1/8 =13/40
So 13/40 of the tank is filled in an hour. We need to find how many hours it will take for the entire tank to be filled. To do so, divide 40 by 13. This gives:
3.08 hours to fill up the tank.
Answer:
a wave passes through a given point or medium what is it called ?
<em><u>Mechanical waves </u></em>