Answer:
a. 0.8 mW/m² b. 1.6 mW/m² c. 6.4 mW/m²
Explanation:
Intensity,I = P/A were P = power and A = area = 4πr² were r = distance from electric guitar = 5.0 m.
a. When P = 0.25 W, I = 0.25 W/4π5² = 0.25 W/100π = 0.00079 W/m² ≅ 0.0008 W/m² = 0.8 mW/m²
b. When P = 0.50 W, I = 0.50 W/4π5² = 0.50 W/100π = 0.00156 W/m² ≅ 0.0016 W/m² = 1.6 mW/m²
c. When P = 2.0 W, I = 2.0 W/4π5² = 2.0 W/100π = 0.00636 W/m² ≅ 0.0064 W/m² = 6.4 mW/m²
Answer:jeff gordon is a boss
Explanation: i am so cool thats how i know its correct
Answer:
A) A warm wire
Explanation:
A warm wire has the most resistance. Heating the metal wire causes atoms to vibrate more, which in turn makes it more difficult for the electrons to flow, increasing resistance. Heating the wire increases resistivity.
Answer:
Option 10. 169.118 J/KgºC
Explanation:
From the question given above, the following data were obtained:
Change in temperature (ΔT) = 20 °C
Heat (Q) absorbed = 1.61 KJ
Mass of metal bar = 476 g
Specific heat capacity (C) of metal bar =?
Next, we shall convert 1.61 KJ to joule (J). This can be obtained as follow:
1 kJ = 1000 J
Therefore,
1.61 KJ = 1.61 KJ × 1000 J / 1 kJ
1.61 KJ = 1610 J
Next, we shall convert 476 g to Kg. This can be obtained as follow:
1000 g = 1 Kg
Therefore,
476 g = 476 g × 1 Kg / 1000 g
476 g = 0.476 Kg
Finally, we shall determine the specific heat capacity of the metal bar. This can be obtained as follow:
Change in temperature (ΔT) = 20 °C
Heat (Q) absorbed = 1610 J
Mass of metal bar = 0.476 Kg
Specific heat capacity (C) of metal bar =?
Q = MCΔT
1610 = 0.476 × C × 20
1610 = 9.52 × C
Divide both side by 9.52
C = 1610 / 9.52
C = 169.118 J/KgºC
Thus, the specific heat capacity of the metal bar is 169.118 J/KgºC
Answer:
8 x 10^-8 C on both the spheres.
Explanation:
Number of electrons added to A = 1 x 10^12
As the sphere A and B are identical to each other so the charge is shared equally.
Charge of one electron = 1.6 x 10^-19 C
Charge on A after wards
= number of electrons after wards x charge of one electron
qA = 0.5 x 10^12 x 1.6 x 10^-19 = 8 x 10^-8 C
Similarly, the charge on sphere B afterwards
= number of electrons after wards x charge of one electron
qB = 0.5 x 10^12 x 1.6 x 10^-19 = 8 x 10^-8 C
