The current in each experiment increases with increase in the voltage. Similarly, the association between resistance and the current in a circuit shows that increase in the resistance shows a reduction in the current, vice versa.
Ohm's Law states that the voltage across an electric conductor is directly proportional to the current(I) passing through it provided the resistant is constant.
So;
V ∝ I
V = IR
where
The objective of this question want us to determine: How did the current change for each test provided that Avery uses a 1.5-volt battery, then she uses a 3-volt battery and lastly she uses a 9-volt battery, given that the resistance is constant through out the whole process.
In the first experiment;
In the second experiment;
In the third experiment;
Therefore, we can conclude that the current in each experiment increases with increase in the voltage. Similarly, the association between resistance and the current in a circuit shows that increase in the resistance shows a reduction in the current, vice versa.
Learn more about Ohm's Law here:
brainly.com/question/14296509
Constant velocity means moving in a straight line at a speed that doesn't change. If the object is moving with constant velocity then its acceleration is zero. Acceleration is the rate at which velocity is changing.
The direction of its displacement wil be
c.northeast
In fact, the dog walks north for 10 meters and east for another 10 meters. The path of the dog can be represented with two vectors, A pointing north (of magnitude 10 meters) and B pointing east (of magnitude 10 meters). The direction of the resultant vector (due to east) will be given by
and the direction will be north-east.
I can think of two of them:
-- carbon monoxide
-- black soot
Answer:
Same magnitude of the 10 nc charge cause the electric field is external.
Explanation:
To do a better explanation, let's go and suppose we have an electric field of, 1300 N/C with a 10 nC charge.
As the system we are talking about is really big, and the charge is small, we can assume always if the charge is sitting right in the same point where the electric field is, then, the electric field would not suffer any kind of alteration in it's value. Therefore, no matter what value of the charge is sitting here, the electric field is independent of the charge, so it would not feel any alteration. However, the force that the charge is feeling would be stronger than in the first case.
F = qE
If charge is doubled, then the force would be bigger in the second case than in the first case, but electric field remain the same value.
