Answer:

Explanation:
The power provided by a resistor (wire in this case) is given by:
.
The resistance of a wire is given by:

Where for the resistivity the one of the copper should be used:
.
The area A is that of a circle, which written in terms of its diameter is:

Putting all together:

Which for our values is:

The answer is Decibels. <span />
Answer:
62.8 μC
Explanation:
Here is the complete question
The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?
Solution
The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ
So, Q = ∫∫∫ρdV
Q = ∫∫∫ρr²sinθdθdrdΦ
Q = ∫∫∫(0.2r²)r²sinθdθdrdΦ
Q = ∫∫∫0.2r⁴sinθdθdrdΦ
We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π
So, Q = ∫∫∫0.2r⁴sinθdθdrdΦ
Q = ∫∫∫0.2r⁴[∫sinθdθ]drdΦ
Q = ∫∫0.2r⁴[-cosθ]drdΦ
Q = ∫∫0.2r⁴-[cosπ - cos0]drdΦ
Q = ∫∫∫0.2r⁴-[-1 - 1]drdΦ
Q = ∫∫0.2r⁴-[- 2]drdΦ
Q = ∫∫0.2r⁴(2)drdΦ
Q = ∫∫0.4r⁴drdΦ
Q = ∫0.4r⁴dr∫dΦ
Q = ∫0.4r⁴dr[Φ]
Q = ∫0.4r⁴dr[2π - 0]
Q = ∫0.4r⁴dr[2π]
Q = ∫0.8πr⁴dr
Q = 0.8π∫r⁴dr
Q = 0.8π[r⁵/5]
Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]
Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]
Q = 0.8π[0.025 m⁵ - 0 m⁵]
Q = 0.8π[0.025 m⁵]
Q = (0.02π mC/m⁵) m⁵
Q = 0.0628 mC
Q = 0.0628 × 10⁻³ C
Q = 62.8 × 10⁻³ × 10⁻³ C
Q = 62.8 × 10⁻⁶ C
Q = 62.8 μC
Answer:
0.546 
Explanation:
From the given information:
The force on a given current-carrying conductor is:

where the length usually in negative (x) direction can be computed as

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:



![F = I (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k](https://tex.z-dn.net/?f=F%20%3D%20I%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7Bx%5E3%7D%7B3%7D%20%5Cbigg%20%5D%20%5E3_1%20%5Chat%20k)
![F = I (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ] \hat k](https://tex.z-dn.net/?f=F%20%3D%20I%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7B3%5E3%7D%7B3%7D%20-%20%5Cdfrac%7B1%5E3%7D%7B3%7D%20%5Cbigg%20%5D%20%20%5Chat%20k)
where;
current I = 7.0 A
![F = (7.0 \ A) (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ] \hat k](https://tex.z-dn.net/?f=F%20%3D%20%287.0%20%5C%20A%29%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7B27%7D%7B3%7D%20-%20%5Cdfrac%7B1%7D%7B3%7D%20%5Cbigg%20%5D%20%20%5Chat%20k)
![F = (7.0 \ A) (9.0) \bigg [\dfrac{26}{3} \bigg ] \hat k](https://tex.z-dn.net/?f=F%20%3D%20%287.0%20%5C%20A%29%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7B26%7D%7B3%7D%20%5Cbigg%20%5D%20%20%5Chat%20k)
F = 546 × 10⁻³ T/mT 
F = 0.546 
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