Answer:
-7.89 * 10^(-9) C
Explanation:
Parameters given:
q1 = 2.42 nC = 2.42 * 10^(-9) C
Distance between q1 and q2 = 5.33 m
q3 = 1.0 nC = 1 * 10^(-9) C
Distance between q1 and q3 = 1.9 m
Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m
The net force acting on q3 is:
F = F(q1, q3) + F(q2, q3)
F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²
F = (9 * 10^(9) * 2.42 * 10^(-9) * 1 * 10^(-9))/3.61 + (9 * 10^(9) * q2 * 1 * 10^(-9))/11.7649
F = 6.033 * 10^(-9) + 0.765*q2
If the net force is zero:
0 = 6.033 * 10^(-9) + 0.765*q2
-0.765*q2 = 6.033 * 10^(-9)
=> q2 = -[6.033 * 10^(-9)]/0.765
q2 = -7.89 * 10^(-9) C
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You’re not completely useless you can always serve as a bad example
Answer:
ΔE> E_minimo
We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference
Explanation:
For this exercise let's use the electric field expression
E = k q / r²
where k is the Coulomb constant that is equal to 9 109 N m² /C², q the charge and r the distance to the point of interest positive test charge, in this case the distance to the bee
let's calculate the field for each charge
Q = 24 pC = 24 10⁻¹² C
E₁ = 9 10⁹ 24 10⁻¹² / 0.20²
E₁ = 5.4 N / C
Q = 32 pC = 32 10⁻¹² C
E₂ = 9 10⁹ 32 10⁻¹² / 0.2²
E₂ = 7.2 N / C
let's find the difference between these two fields
ΔE = E₂ -E₁
ΔE = 7.2 - 5.4
ΔE = 1.8 N / C
the minimum detection field is
E_minimum = 0.77 N / C
ΔE> E_minimo
We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference
Answer:
please give me brainlist and follow
Explanation:
Gravitational force keeps it attracted to the Earth, and centripetal force keeps it moving in a circle. ... Explain how Newton's third law and his law of universal gravitation are connected. His third law states that every force has an equal opposite force attracted to it, and that force is caused by gravity.
Answer:
9.8 m/s/s
Explanation:
The numerical value, in meters per second squared, of the acceleration of an object experiencing true free fall is 9.8 m/s/s. This is called the acceleration due to gravity.
