Answer:
1) Current decreases; 2) Inverse proportionally; 3) 1[A]
Explanation:
1)
As we can see as the resistance increases the current decreases, if we take two points as an example, when the resistance is equal to 50 [ohms] the current is equal to 1[amp] and when the resistance is equal to 200 [ohms] the current tends to have a value below 0.5 [amp]. Thus demonstrating the decrease in current.
2)
Inverse proportionally, by definition we know that the law of ohm determines the voltage according to resistance and amperage. This is the voltage will be equal to the product of the voltage by the resistance.
![V=I*R\\V = voltage [volts]\\I = current[amp]\\R = resistance [ohms]](https://tex.z-dn.net/?f=V%3DI%2AR%5C%5CV%20%3D%20voltage%20%5Bvolts%5D%5C%5CI%20%3D%20current%5Bamp%5D%5C%5CR%20%3D%20resistance%20%5Bohms%5D)
where:
And whenever we have in a fractional number the denominator the variable we are interested in, we can say that this is inversely proportional to the value we are interested in determining. In this case, we can see from the two previous expressions that both the current and the resistance appear in the denominator, therefore they are inversely proportional to each other.
3)
If we place ourselves on the graph on the resistance axis, we see that at 50 [ohm] will correspond a current value equal to 1 [A].
A) See ray diagram in attachment (-6.0 cm)
By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:
where
q is the distance of the image from the lens
f = -10 cm is the focal length (negative for a diverging lens)
p = 15 cm is the distance of the object from the lens
Solving for q,
B) The image is upright
As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:
where 
are the size of the image and of the object, respectively.
Since q < 0 and p > o, we have that 
, which means that the image is upright.
C) The image is virtual
As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.
This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual
P = V^2 / R.
So, 3.3^2 / 0.025 = 435.6W.
Note, you can get the power equation from:
P = V*I. Also, I = V/R.
Substituting V/R in for I in the 1st equation, you get P = V^2 / R.
Answer:
180^2 + 390^2 = force ^2 (Pythagoras) root of force^2 = 429.5N approx resultant force Acceleration = Force/Mass 429.5/270 = 1.5907 ms^-2 in a Southwesterly direction.
Explanation:
