Answer: a) 335.8 μm; b) keeping the same radius, the new has double potential, V=1340V so if teh radius is also double the potentail is the same (V=670V).
Explanation: In order to explain this problem we have to consider the potential given for sphere respect to infinity ( V=0) in the form:
V=k*Q/R the we have
R=k*Q/V= 9*10^9*25*10^-12/670=335.8 *10^-6 m
When two drop join to form a single drop (considering with the same radius) we have:
V=k*2Q/R
So the new V is double the original,
V=9*10^9*2*25*10^-12/335.8*10^-6=1340V
if the final single drop has a 2R of radius so
V=k*2Q/2R= 670 V
It has the same original potential.
Answer:
i think b
Explanation:because they are so close
Answer:
The weight of the probe is 50 Newtons
Explanation:
Newtons second law states that F = ma
Given the mass of 25kg, and the acceleration of 2m/s^2, we can substitute both values into the equation to find the weight force.
The weight of the probe is 50 Newtons
Answer:
What is the pressure transmitted in the liquid on a hydraulic pump where an elephant with a weight of 40 000 N is placed on top of the large piston with an area of 40 m2. The small piston area is 4 m2.
Answer:
The answer is "The last choice".
Explanation:
Please find the complete question in the attachment.
In an external electric field, its electrical energy at positive charge becomes directed to just the electrical domain. Therefore it will speed towards its base plate whenever cyber one is released to rest at h0. It was never going to reach the top plate. Thus, the last choice corrects because in this the cyber-on never reaches its upper stage.