<span>Place a test charge in the middle. It is 2cm away from each charge.
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point.
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out.
THIS IS A TRICK QUESTION.
THe electric field exactly midway between them = 0/Q = 0.
But if the point moves even slightly you need the following formula
F= (1/4Piε)(Q1Q2/D^2)
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)</span>
force between two parallel wire is
f/l = mueo*i1i2/2pir
f/l = 2*10^-7*i1i2/rl
i2 = f*r/2*10^-7*i1
i2 = 342.9 A
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Answer: 477W
Explanation:
Given the following :
Mass (m) = 7.3kg
Initial Velocity (u) = 0
Final velocity (v) = 14m/s
time (t) = 1.5s
Power = workdone (W) / time (t)
The workdone can be calculated as the change in kinetic energy (KE) :
Recall ;
KE = 0.5mv^2
Therefore, change in KE is given by:
0.5mv^2 - 0.5mu^2
Change in KE = 0.5(7.3)(14^2) - 0.5(7.3)(0^2)
Change in KE = 715.4J
Therefore ;
Average power = Workdone / time
Workdone = change in KE = 715.4N
Average power = 715.4 / 1.5
Average power = 476.93333 W
= 477W
Star 1 - 4 hours right ascension
Star 2 - 3 hours right ascension
Subtracting hours right ascension
4 hours right ascension - 3 hours right ascension = 1 hours right ascension.
Thus,
1) star 1 will rise 1 hour before star 2
