The net force on q₃ will be 17.51 N. The net force is the algebraic sum of the two forces on the pleading q₃
<h3 /><h3>What is Columb's law?</h3>
The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
The force,by the charge q₁ on the q₃;
The force,by the charge q₂ on the q₃;
The net force is the sum of the two forces;
Hence, the net force on q₃ will be 17.51 N.
To learn more about Columb's law, refer to the link;
brainly.com/question/1616890
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A 50kg object on earth weighs 81.67 on the moon
Answer:
Friction does not oppose motion, it oppose relative motion.
Explanation:
static friction opposes relative motion
may i ask your answer soloutions... and i believe that the correct answer is mass
The answer is C.
The question says the potential difference is what is changing, which means we're solving for V.
It tells us that potential difference increases by a factor of two, which just means V doubles.
With this info, we can pick some numbers, plug it into Ohms law and see what happens.
Here's an example where I just picked random numbers that are easy to work with:
V=I*R
10=I*5
I=2
Lets increase the potential difference (V) by a factor of two and see what happens to current:
V=I*R
20=I*5 (all I've done is double the potential difference from 10 to 20)
I=4
When we increase V by a factor of 2, I increases by a factor of 2. We went from I=2 to I=4.
We can increase V by factor of 2 again and see:
V=I*R
40=I*5
I=8
Okay, current just increased by a factor of 2 again when we increased the potential difference by a factor of 2.
It's always good to check work with alternate numbers, so here's one more set:
V=I*R
16=I*4
(remember, we know we're solving for V, so I'm just plugging in random numbers for I and R)
I=4
Increase V by factor of 2:
32=I*4
I=8
So, when we increase V (the potential difference) by a factor of 2, I (current) always increases by a factor of 2 as well.
Hope this helps!