Answer:
Talk to his team captain and ask him to alert the referee to keep a better eye on the player.
D = 497.4x10⁻⁶m. The diameter of a mile of 24-gauge copper wire with resistance of 0.14 kΩ and resistivity of copper 1.7×10−8Ω⋅m is 497.4x10⁻⁶m.
In order to solve this problem we have to use the equation that relates resistance and resistivity:
R = ρL/A
Where ρ is the resistivity of the matter, the length of the wire, and A the area of the cross section of the wire.
If a mile of 24-gauge copper wire has a resistance of 0.14 kΩ and the resistivity of copper is 1.7×10⁻⁸ Ω⋅m. Determine the diameter of the wire.
First, we have to clear A from the equation R = ρL/A:
A = ρL/R
Substituting the values
A = [(1.7×10⁻⁸Ω⋅m)(1.6x10³m)]/(0.14x10³Ω)
A = 1.9x10⁻⁷m²
The area of a circle is given by A = πr² = π(D/2)² = πD²/4, to calculate the diameter D we have to clear D from the equation:
D = √4A/π
Substituting the value of A:
D = √4(1.9x10⁻⁷m²)/π
D = 497.4x10⁻⁶m
<span>They deliver oxygen to the blood.
For the regular body activities to be carried out.
Acid-base balance or body temperature control.</span>
Answer:
![P_1 = P_2 = \frac{P}{2}](https://tex.z-dn.net/?f=P_1%20%3D%20P_2%20%3D%20%5Cfrac%7BP%7D%7B2%7D)
so each bulb brightness becomes half of its given or indicated power
![P_1 = P_2 = \frac{V^2}{R}](https://tex.z-dn.net/?f=P_1%20%3D%20P_2%20%3D%20%5Cfrac%7BV%5E2%7D%7BR%7D)
so both bulb will glow same power as indicated
Explanation:
Let the indicated power on the bulbs is given as P and its rated voltage is V
so here resistance of each bulb is given as
![R = \frac{V^2}{P}](https://tex.z-dn.net/?f=R%20%3D%20%5Cfrac%7BV%5E2%7D%7BP%7D)
now if the two bulbs are connected in series so we will have
![R_{eq} = R_1 + R_2](https://tex.z-dn.net/?f=R_%7Beq%7D%20%3D%20R_1%20%2B%20R_2)
![R_{eq} = 2\frac{V^2}{P}](https://tex.z-dn.net/?f=R_%7Beq%7D%20%3D%202%5Cfrac%7BV%5E2%7D%7BP%7D)
now the current in the circuit is given as
![i = \frac{V}{R_{eq}](https://tex.z-dn.net/?f=i%20%3D%20%5Cfrac%7BV%7D%7BR_%7Beq%7D)
![i = \frac{P}{2V}](https://tex.z-dn.net/?f=i%20%3D%20%5Cfrac%7BP%7D%7B2V%7D)
now brightness of each bulb is given as
![P_1 = P_2 = i^2 R](https://tex.z-dn.net/?f=P_1%20%3D%20P_2%20%3D%20i%5E2%20R)
![P_1 = P_2 = \frac{P}{2}](https://tex.z-dn.net/?f=P_1%20%3D%20P_2%20%3D%20%5Cfrac%7BP%7D%7B2%7D)
so each bulb brightness becomes half of its given or indicated power
Now if the two bulbs are connected in parallel
then the net voltage across each bulb is "V"
so we will have
![P_1 = P_2 = \frac{V^2}{R}](https://tex.z-dn.net/?f=P_1%20%3D%20P_2%20%3D%20%5Cfrac%7BV%5E2%7D%7BR%7D)
so both bulb will glow same power as indicated
Answer: THE ANSWER IS C!
The total energy of a system can decrease only if energy leaves the system.
Explanation: Apex!