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].
25m north east hope this helps
Answer:
<h2>
650W/m²</h2>
Explanation:
Intensity of the sunlight is expressed as I = Power/cross sectional area. It is measured in W/m²
Given parameters
Power rating = 6.50Watts
Cross sectional area = 100cm²
Before we calculate the intensity, we need to convert the area to m² first.
100cm² = 10cm * 10cm
SInce 100cm = 1m
10cm = (10/100)m
10cm = 0.1m
100cm² = 0.1m * 0.1m = 0.01m²
Area (in m²) = 0.01m²
Required
Intensity of the sunlight I
I = P/A
I = 6.5/0.01
I = 650W/m²
Hence, the intensity of the sunlight in W/m² is 650W/m²
Well let's see:
A). No. A capacitor doesn't measure anything.
B). No. The power delivered to the circuit is determined by
the battery or power supply and all the things in
the circuit that dissipate energy. A capacitor doesn't
do any of these things.
C). No. If any current actually flows between its plates,
the capacitor is shot and can't do its job, and
must be replaced.
D). Yes. A capacitor stores charges on its plates, and
electrical energy in the field between its plates.
Answer:
1,1 m
Explanation:
Dado que;
coeficiente de fricción = 0,6
sabemos que W = R = mgcos 37 = 3.5Kg * 10m / s ^ 2 * cos37 = 27.95 N
coeficiente de fricción = fuerza / reacción normal (R)
Fuerza = 0.6 * 27.95 N
Fuerza (F) = 16.77 N
Recuerda que F = Ke
dónde;
K = constante de fuerza (15N / m)
e = extensión (lo desconocido)
e = F / K
e = 16,77 N / 15 N / m
e = 1,1 m
