Physics, 18.10.2019 04:40, DnakMeme6e7272727272

The 10 ω resistor in the figure is dissipating 40 w of power. (figure 1) how much power is the 20ω resistor dissipating?

The 10 ω resistor in the figure is dissipating 40 w of power. (figure 1) how much power is the 20ω r

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answered: pickelswolf3036

20W

Explanation:

The first thing to bear in mind is that the 10 ohms and 20 ohms resistors are in parallel; and when resistors are in parallel the voltage drop across them will be the same.

Therefore if we can obtain the voltage drop across the 10 ohms resistor, then we can apply the same voltage to the 20 ohms resistor (since it is in parallel with the 10 ohms resistor) and then use that same voltage to obtain the power dissipated therein.

The following are the relationships between power (p), current (I), resistance (R) and voltage (V);

$V=IR.............(1)\\P=IV..............(2)\\$

Substituting equation(1) into (2), we obtain the following;

$P=I(IR)\\P=I^2R............(3)$

Let the power across the 10 ohms resistor be $P_{1$ and that of the 20 ohms resistor be $P_{2$ . Also their their respective currents and voltages be I_1 , I_2 , V_1 and V_2 .

Given;

R_1=10ohms; P_1=40W ; R_2=20ohms; P_2=?

Using equation (3), we can obtain I_1 as follows;

$P_1=I_1^2R_1\\40=I_1^2*10\\therfore\\I_1^2=40/10=4\\I_1=\sqrt{4}=2A$

We now use this to obtain the voltage drop across R_1 as follows;

$V_1=I_1R_1\\V_1=2*10=20V$

Recall we said that the voltage drop across the 10 ohms and the 20 ohms resistors are the same because they are in parallel, therefore V_1=V_2=20V .

We now use this value to determine the current in R_2 As follows through the relationship defined in equation(1);

$V_2=I_2R_2\\therefore\\I_2=V_2/R_2\\I_2=20/20=1A$

Finally we now use equation (2) to determine P_2 as follows

$P_2=I_2V_2\\P_2=1*20\\P_2=20W$

Note please: This is not the only method that could be used in solving this problem. There are other approaches that could be used, but I feel the approach used here should be simplified and straight forward to understand. Thanks.

answered: Guest

answer;

d. are made mostly of liquid

explanation;

-the general composition and structure planets includes; they have a thick gas and liquid layer covering a small,solid core. mostly the gases are hydrogen and helium atmosphere and layers of liquid hydrogen. for instance neptune and uranus are examples of the outer planets, they have a core covered by ice and ammonia.

-neptune's atmosphere is hydrogen and helium with a bit of methane so there is no oxygen to breath and the interior is made up of frozen water and ammonia with a rock and iron core. thus, there is no liquid water or solid surface to survive.

answered: Guest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .