<h3>
</h3><h3>Given</h3>
v = 20m\s
a = 3m\s^2
t = 4sec
Firstly we have to find u
a = 
3m\s =
12m\s = 20 - u
20 - u = 12m\s
- u = -8
u = 8
Now we can easily find distance by using second equation of motion
s = ut + 1\2 at^2
s = 8(4) + 1\2(3)(16)
s = 32 + 24
s = 56
So distance is 56 m\s hope it helps
Answer:
The answers are in the explanation section below
Explanation:
1) The generalization that can be made from the exploration is that as we move away from the positive electrode, the potential energy gets lower. If we move away from the negative electrode, then the potential energy becomes higher.
2) The positive test charge will have the least potential energy when it gets to the negative electrode point.
3) To move one electron 1m in a direction along one of the equal potential lines, there is no energy needed. Zero work will be required for a charge to move on the equipotential line.
4) If lightning strikes a tree 20m away, it would be better to face the tree or have our back facing the tree. This is because the equipotential line will be present at the point where our body stands, this will protect from electric shock.
The pattern to be sketched is attached.
Answer:
Explanation:
The resistance of a wire is given by:
where
is the resistivity of the material
L is the length of the wire
A is the cross-sectional area of the wire
1) The first wire has length L and cross-sectional area A. So, its resistance is:
2) The second wire has length twice the first one: 2L, and same thickness, A. So its resistance is
3) The third wire has length L (as the first one), but twice cross sectional area, 2A. So, its resistance is
By comparing the three expressions, we find
So, this is the ranking of the wire from most current (least resistance) to least current (most resistance).
Then the tangent of angle-Θ is (Ay / Ax).
