Answer:
1. B. Increase
2. A. Decrease
Explanation:
To understand this issue, we need to put some values and using the ohm's law we can corroborate the two situations.
Ohm's law:
![V = I*R\\where:\\I = current[A] ampers\\R = resistance [ohms]\\V = voltage [volts]\\](https://tex.z-dn.net/?f=V%20%3D%20I%2AR%5C%5Cwhere%3A%5C%5CI%20%3D%20current%5BA%5D%20ampers%5C%5CR%20%3D%20resistance%20%5Bohms%5D%5C%5CV%20%3D%20voltage%20%5Bvolts%5D%5C%5C)
Now for the voltage we will use V = 110 [V], for resistance R = 10 [ohms]
Replacing the values we have:
![I = \frac{V}{R} \\\\I = \frac{110}{10}\\I= 11 [amp]](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BV%7D%7BR%7D%20%5C%5C%5C%5CI%20%3D%20%5Cfrac%7B110%7D%7B10%7D%5C%5CI%3D%2011%20%5Bamp%5D)
Now let's double the voltage 220 [V]:
![I = \frac{220}{10} \\I = 22 [amp]\\](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B220%7D%7B10%7D%20%5C%5CI%20%3D%2022%20%5Bamp%5D%5C%5C)
Therefore the current will be increased.
Let's do the same for the resistance if originally we have R = 10 [ohms]
![I = \frac{V}{R}\\I = \frac{110}{10} \\I = 11 [amp]\\](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BV%7D%7BR%7D%5C%5CI%20%3D%20%5Cfrac%7B110%7D%7B10%7D%20%5C%5CI%20%3D%2011%20%5Bamp%5D%5C%5C)
Now let's double the resistance 20 [ohms]:
![I=\frac{110}{20}\\I = 5.5 [amp]](https://tex.z-dn.net/?f=I%3D%5Cfrac%7B110%7D%7B20%7D%5C%5CI%20%3D%205.5%20%5Bamp%5D)
Therefore the current will be decreased.
Technically, this is true...
I'm a bit rusty at this, tho
<span>The ball with an initial velocity of 2 m/s rebounds at 3.6 m/s
The ball with an initial velocity of 3.6 m/s rebounds at 2 m/s
There are two principles involved here
Conservation of momentum and conservation of energy.
I'll use the following variables
a0, a1 = velocity of ball a (before and after collision)
b0, b1 = velocity of ball b (before and after collision)
m = mass of each ball.
For conservation of momentum, we can create this equation:
m*a0 + m*b0 = m*a1 + m*b1
divide both sides by m and we get:
a0 + b0 = a1 + b1
For conservation of energy, we can create this equation:
0.5m(a0)^2 + 0.5m(b0)^2 = 0.5m(a1)^2 + 0.5m(b1)^2
Once again, divide both sides by 0.5m to simplify
a0^2 + b0^2 = a1^2 + b1^2
Now let's get rid of a0 and b0 by assigned their initial values. a0 will be 2, and b0 will be -3.6 since it's moving in the opposite direction.
a0 + b0 = a1 + b1
2 - 3.6 = a1 + b1
-1.6 = a1 + b1
a1 + b1 = -1.6
a0^2 + b0^2 = a1^2 + b1^2
2^2 + -3.6^2 = a1^2 + b1^2
4 + 12.96 = a1^2 + b1^2
16.96 = a1^2 + b1^2
a1^2 + b1^2 = 16.96
The equation a1^2 + b1^2 = 16.96 describes a circle centered at the origin with a radius of sqrt(16.96). The equation a1 + b1 = -1.6 describes a line with slope -1 that intersects the circle at two points. Those points being (a1,b1) = (-3.6, 2) or (2, -3.6). This is not a surprise given the conservation of energy and momentum. We can't use the solution of (2, -3.6) since those were the initial values and that would imply the 2 billiard balls passing through each other which is physically impossible. So the correct solution is (-3.6, 2) which indicates that the ball going 2 m/s initially rebounds in the opposite direction at 3.6 m/s and the ball originally going 3.6 m/s rebounds in the opposite direction at 2 m/s.</span>
Answer:
it functions based on rotatry motion..
Explanation:
because it is stationary....and has the ability to hold things in it...in a kind of random motion
