Answer:
Option E
Explanation:
In the presence of two point charges at the two vertices of an equilateral triangle, the resultant electric field at the third vertex due to these charges can not be zero whether the charges are identical or not.
The reason being that only of the x or y component of the field can be cancelled out in either case still the total field can't be reduced to zero.
This can only be achieved if another charge is present.
Answer:
The force on q₁ due to q₂ is (0.00973i + 0.02798j) N
Explanation:
F₂₁ = ![\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|}](https://tex.z-dn.net/?f=%5Cfrac%7BK%7Cq_1%7Cq_2%7C%7D%7Br%5E2%7D.%5Cfrac%7Br_2_1%7D%7B%7Cr_2_1%7C%7D)
Where;
F₂₁ is the vector force on q₁ due to q₂
K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²
r₂₁ is the unit vector
|r₂₁| is the magnitude of the unit vector
|q₁| is the absolute charge on point charge one
|q₂| is the absolute charge on point charge two
r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)
|r₂₁| = ![\sqrt{(4^2)+(11.5^2)} = \sqrt{148.25}](https://tex.z-dn.net/?f=%5Csqrt%7B%284%5E2%29%2B%2811.5%5E2%29%7D%20%3D%20%5Csqrt%7B148.25%7D)
(|r₂₁|)² = 148.25
![F_2_1=\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|} = \frac{8.99X10^9(14X10^{-6})(60X10^{-6})}{148.25}.\frac{(4i + 11.5j)}{\sqrt{148.25} }](https://tex.z-dn.net/?f=F_2_1%3D%5Cfrac%7BK%7Cq_1%7Cq_2%7C%7D%7Br%5E2%7D.%5Cfrac%7Br_2_1%7D%7B%7Cr_2_1%7C%7D%20%3D%20%5Cfrac%7B8.99X10%5E9%2814X10%5E%7B-6%7D%29%2860X10%5E%7B-6%7D%29%7D%7B148.25%7D.%5Cfrac%7B%284i%20%2B%2011.5j%29%7D%7B%5Csqrt%7B148.25%7D%20%7D)
= 0.050938(0.19107i + 0.54933j) N
= (0.00973i + 0.02798j) N
Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N
What happens when you pour water in a glass? It takes the shape of the glass. This means that water can't have a fixed shape or volume
The molecular weight of carbon dioxide (CO2) is 44.00 a.m.u., and the molecular weight of propane gas (C3H8) is 44.10 a.m.u. Thus CO2 diffuses_______ C3H8.
Slower Than
Faster Than
The Same As
Slower than
Answer:
Answer is A
Explanation:
As we know , for constant velocity we get a straight line.
The formula for this problem is s = vt which is similar to a straight line formula like y = mx + c.
If we put here c = 0 we get the formula for distance and velocity.
So the answer is A.