Answer:
The magnitude of electric field is 22.58 N/C
Solution:
Given:
Force exerted in upward direction,
Charge, Q =
Now, we know by Coulomb's law,
Also,
Electric field,
Thus from these two relations, we can deduce:
F = QE
Therefore, in the question:
Here, the negative side is indicative of the Electric field acting in the opposite direction, i.e., downward direction.
The magnitude of the electric field is:
Answer:
the final velocity of the two blocks is
the distance that A slides relative to B is
Explanation:
From the diagram below;
acceleration of A relative to B is :
where
v = u + at
Making t the subject of the formula; we have:
which implies the distance that A slides relative to B.
The final velocities of the two blocks can be determined as follows:
v = u + at
Thus, the final velocity of the two blocks is
Answer:
The magnitude and direction of electric field midway between these two charges is along AB.
Explanation:
Given that,
First charge
second charge
Distance = 20 cm
We need to calculate the electric field
For first charge,
Using formula of electric field
Put the valueinto the formula
Direction of electric field along AB
We need to calculate the electric field
For second charge,
Using formula of electric field
Put the valueinto the formula
Direction of electric field along AO
We need to calculate the net electric field at midpoint
Direction of net electric field along AB
Hence, The magnitude and direction of electric field midway between these two charges is along AB.
This is called vaporization.
hope this helps :)
The car heads east at an average speed of 50 miles per hour from the intersection point towards East. The truck heads east at an average speed of 60 miles per hour from the intersection point towards South.
The distance of car from the intersection point after t hours is .
The distance of truck from the intersection point after t hours is .
Since these distances are perpendicular to each other, distance apart d (in miles) at the end of t hours is
Thus the distance apart is