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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Answer: p= m/v so 90kg/.075m^3 = 1,200
2a. .35 m 1.1 m and .015 m
2b. 35 cm x 110 cm x 1.5 cm = 5,775 cm^3 = 57.75 m^3
mass= pv
2700•57.75= 155,925 kg
mass= 155,925 kg
volume= 57.75 m^3
Explanation: physics
This problem involves Newton's universal law of gravitation and the equation to follow would be.
F = GM₁M₂/r²
Given: M₁ = 0.890 Kg; M₂ = 0.890 Kg; F = 8.06 x 10⁻¹¹ N; G = 6.673 X 10⁻¹¹ N m²/Kg²
Solving for distance r = ?
r = √GM₁M₂/F
r = √(6.673 x 10⁻¹¹ N m₂/Kg²)(0.890 Kg)(0.890 Kg)/ 8.06 x 10⁻¹¹ N
r = 0.81 m
Resultant force= (2*6^2)^(1/2)
=8.5m/s
answer is B.