For an object moving with constant negative acceleration, draw the following:
1 answer:
Answer:
Kindly find the graphs attached
Explanation:
For figure 1: There is a steady increase in the position of the object as time increases. This is because despite the negative acceleration (deceleration), the object continues to move and cover more ground as time goes by.
<em>The straight line graph is observed because the acceleration is constant and not varying.</em>
For Figure 2: The graph of velocity vs time will have an inverted nature. This is because since the object is decelerating, it is reducing in its velocity as time goes by (increases). <em>This is also in a straight line since the deceleration is constant.</em>
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If the distance between two charges is halved, the electrical force between them increases by a factor 4.
In fact, the magnitude of the electric force between two charges is given by:
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges
We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
the magnitude of the force changes as follows:
so, the force increases by a factor 4.
In fact, the magnitude of the electric force between two charges is given by:
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges
We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
the magnitude of the force changes as follows:
so, the force increases by a factor 4.
The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.
The relationship between the resistance and the resistivity of a wire is
where
is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length
the cross-sectional area is given by
where r is the radius of the wire. Substituting in the previous equation ,we find
For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
Therefore, the new resistivity must be 4 times the original one.
The relationship between the resistance and the resistivity of a wire is
where
A is the cross-sectional area
R is the resistance
L is the wire length
the cross-sectional area is given by
where r is the radius of the wire. Substituting in the previous equation ,we find
For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
Therefore, the new resistivity must be 4 times the original one.