Remark
When you are asked a question like this, the first thing to do is search out a formula and put some limits on it.
Formula
I = E/R which comes from E = IR. To get to the derived formula, divide both sides by R
E/R = I*R/R
E/R = I
Discussion
This is an inverse relationship. That means that as one goes up the other one will go down.
So in this case you keep E constant and you manipulate R and look at your results for I
Case 1
Let us say that E = 10 volts
Let us also say the R = 10 ohms
I = E/R
I = 10/10
I = 1 ohm
Case Two
Let's raise the Resistance to 100 ohms
E = 10
R = 100
I = 10/100 = 0.1
Conclusion
As the Resistance goes up, the current goes down. Answer: A
- Initial velocity (u) = 0 m/s [the car was at rest]
- Distance (s) = 80 m
- Time (t) = 10 s
- Let the magnitude of acceleration be a.
- By using the equation of motion,
we get,
<u>A</u><u>nswer:</u>
<u>The </u><u>magnitude</u><u> </u><u>of </u><u>its </u><u>acceleration</u><u> </u><u>is </u><u>1</u><u>.</u><u>6</u><u> </u><u>m/</u><u>s^</u><u>2</u><u>.</u>
Hope you could get an idea from here.
Doubt clarification - use comment section.
Answer:
Explanation:
This is a simple gravitational force problem using the equation:
where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.
Filling in:
I'm going to do the math on the top and then on the bottom and divide at the end.
and now when I divide I will express my answer to the correct number of sig dig's:
6.45 × 10¹⁶ N
Answer:
E = 2k 
Explanation:
Gauss's law states that the electric flux equals the wax charge between the dielectric permeability.
We must define a Gaussian surface that takes advantage of the symmetry of the problem, let's use a cylinder with the faces perpendicular to the line of charge. Therefore the angle between the cylinder side area has the same direction of the electric field which is radial.
Ф = ∫ E . dA = E ∫ dA = q_{int} /ε₀
tells us that the linear charge density is
λ = q_ {int} /l
q_ {int} = l λ
we substitute
E A = l λ /ε₀
is area of cylinder is
A = 2π r l
we substitute
E =
E =
the amount
k = 1 / 4πε₀
E = 2k 