Hello!
Since the two weights are <em>off</em> the table, the block will move towards letter F.
I hope this helps :))
<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:
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Answer:
437 Joules
Explanation:
Use the formula for work directly
(work) = (force) x (displacement)
to get
(work) = (19 N) x (23 m) = 437 Joules
Given :
A box weighing 12,000 N is parked on a 36° slope.
To Find :
What will be the component of the weight parallel to the plane that balances friction.
Solution :
The component of that will be parallel to the plane to balance friction is :
Therefore, component of force to balance friction is F sin 36° .
Hence, this is the required solution.
