At rest because if the distance is not changing, then it is not moving any further, so it must not be moving! The time keeps going no matter what, so the distance, whether it is 0 m or 10,000 km, if the y is horizontal the distance does not change.
Answer:
<em>To reverse the direction of an electric current, we simply reverse the voltage either automatically with the help of some switching circuitry or manually by changing the voltage source terminals connection. </em>
Explanation:
For electric current to flow, there must be a potential difference, usually referred to as the voltage. The electric current flow is analogous to the flow of water under the action of a pump, through a series of pipe connections. The voltage is similar to the driving action of the pump, and current flows the same way water flows. The resistance due to drag on the pipe wall is equivalent to electric resistance. For current to flow in the reverse direction, the voltage or rather, the potential difference is changed, causing the current to flow in the opposite direction. This can be done by switching the terminals of the voltage source, or by automatic means. The automatic switching can be done with a transistor based circuitry.
The magnitude of the vector C is 96.32m
<h3>How to solve for the magnitude of vector c</h3>
Ax = AcosθA
= 40 cOS 20
= 37.59
Ay = AsinθA
-40sin20
= -13.68
Bx = B cos θ B
= 75Cos50
= 48.21
By = BsinθB
= 75sin50
= 57.45
Cx = AX + Bx
= 37.59 + 48.21
= 85.8
Cy = Ay + By
= -13.65 + 57.45
= 43.77
The magnitude is solved by
|c| = ![\sqrt{Cx^{2}+Cy^{2} }](https://tex.z-dn.net/?f=%5Csqrt%7BCx%5E%7B2%7D%2BCy%5E%7B2%7D%20%20%7D)
= √85.8² + 43.77²
= 96.32m
The magnitude of the vector c is 96.32m
Read more on the magnitude of a vector here:
brainly.com/question/3184914
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I'm actually going ahead in the book (DC Circuits) so this isn't really homework but I figured the tag was appropriate....the name of the chapter is Ohm's Law and Watt's Law.
<span>Problem: Calculate the power dissipated in the load resistor, R, for each of the circuits.Circuit (a): V = 10V; I = 100mA; R = ?; Since I know
V and
I use formula
P = IV: P = IV = (100mA)(10V) = 1 W.</span>
The next question is what I'm not sure about:
Question: What is the power in the circuit (a) above if the voltage is doubled? (Hint: Consider the effect on current).
What I did initially was: P = IV = (100mA)(2V) = 2 W
But then I looked at the answer and it said 4 W, then I looked at the Hint again. Then I remembered in the book early on it said "If the voltage increases across a resistor, current will increase."
So question is: When solving problems I have to increase (or decrease) current (I) every time voltage (V) is increased (decreased) in a problem, right? How about the other way around, when increasing current (I), you need to increase voltage (V). I'm pretty sure that's how they got 4 W, but want to make sure before I head to the next section of the book.
P = IV = (200mA)(2V) = 4 W
Answer:
Explanation:
Polar coordinates formula
Summary. To convert from Polar Coordinates (r,θ) to Cartesian Coordinates (x,y) : x = r × cos( θ ) y = r × sin( θ )