The Electric field is zero at a distance 2.492 cm from the origin.
Let A be point where the charge
C is placed which is the origin.
Let B be the point where the charge
C is placed. Given that B is at a distance 1 cm from the origin.
Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.
i.e., at distance 'x' from B.
Using Coulomb's law,
,
= 



k is the Coulomb's law constant.
On substituting the values into the above equation, we get,

Taking square roots on both sides and simplifying and solving for x, we get,
1.67x = 1+x
Therefore, x = 1.492 cm
Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.
Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926
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Because gravity is constant
<span>the only force acting in free-fall is gravity which points downward at 9.8 m/s</span>
Answer:
1.7323
Explanation:
To develop this problem, it is necessary to apply the concepts related to refractive indices and Snell's law.
From the data given we have to:



Where n means the index of refraction.
We need to calculate the index of refraction of the liquid, then applying Snell's law we have:



Replacing the values we have:


Therefore the refractive index for the liquid is 1.7323
Options:
universal appeal
flattery
association
bandwagon
Answer: Association
Explanation: Advertisement is a marketing technique through which business organisations utilize the opportunities provided by both the print, electronic and other channels of communication to market a product to the target audience.
Association Advertising is a type of Advertising where certain attributes which have been known to be associated with good and quality products are used to market the products to the target audience.
Answer:
d
Explanation:
In physics and engineering, a free body diagram (force diagram, or FBD) is a graphical illustration used to visualize the applied forces, moments, and resulting reactions on a body in a given condition.