<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>
Answer:
C. Plant A orbits its star faster than Plant B
Explanation:
Did it on study island
The distance travelled by the ball that is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s is 5.34 m
s = ut + 1 / 2 at²
s = Distance
u = Initial velocity
t = Time
a = Acceleration
Vertically,
s = 15.4 m
u = 0
a = 9.8 m / s²
15.4 = 0 + ( 1 / 2 * 9.8 * t² )
t² = 3.14
t = 1.77 s
Horizontally,
u = 3.01 m / s
a = 0 ( Since there is no external force )
s = ( 3.01 * 1.77 ) + 0
s = 5.34 m
Therefore, the distance travelled by the ball before hitting the ground is 5.34 m
To know more about distance travelled
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Answer: 31.6ft
Explanation:
Check the attachment for the diagram.
According to the right angle triangle AEC, we will use Pythagoras theorem to get |AC|. Note that |AE| = |AB| - |CD|
that is 20ft - 10ft = 10ft
According to the theorem, the square of the sum of the adjacent side and the opposite side is equal to the square of the hypotenuse.
|AE|^2 + |EC|^2 = |AC|^2
10^2 + 30^2 = |AC|^2
100 + 900 = |AC|^2
|AC| = √1000
|AC| = 31.6ft
Therefore, the wire should be anchored 31.6ft to the ground to minimize the amount of wire needed.
Answer:
Explanation:
Given that,
The current flowing in the circuit, I = 3 A
The power of the battery, P = 25 W
We need to find the resistance of the battery. We know that the power of the battery is given by the formula as follows :
Put all the values to find R.
So, the resistance is equal to 
.
