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Artist 52 [7]
3 years ago
11

7. A cube sits in the Cartesian coordinate system. Each side of the cubeis 2m. Electric charge is distributed in the cube with a

density functionV=xy2e2z(C=m3). Find the total charge in the cube
Physics
1 answer:
Zanzabum3 years ago
4 0

Answer:

Net charge inside the cube is 321.6 C

Explanation:

Given charge density function is ρ = xy^2 e^{2z}

side of cube is 2m

According to the definition of volume charge density total charge is given by

q = total charge = \int {\rho} \, dv where dv = dxdydz is the volume element

on substututing the respected values we get

q = \int\limits^2_0 {x} \, dx \int\limits^2_0 {y^2} \, dy \int\limits^2_0 {e^2^z} \, dz  

on solving the above integration we get

q =  \frac{x^2}{2} x \frac{y^3}{3} x \frac{e^2^z}{2}

⇒ q = 321.6 C

Therefore net charge inside the cube is 321.6 C

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3 years ago
What mass, in grams, of aluminum fins could 2571 J of energy heat from 15.73 ∘C to 26.50 ∘C ? Aluminum has a specific heat of 0.
nadya68 [22]

Answer:

266 g or 0.266 kg

Explanation:

The formula for specific heat capacity is given as,

Q = cm(t₂-t₁) ..................... Equation 1

Where Q = Heat Energy, c = specific heat capacity of Aluminum, m = mass of the aluminum fins, t₁ = initial temperature, t₂ = final temperature.

make m the subject of the equation,

m = Q/c(t₂-t₁)................... Equation 2

Given : Q = 2571 J, c = 0.897 J/g.°C, t₁ = 15.73 °C, t₂ = 26.50 °C.

Substitute into equation 2

m = 2571/[0.897×(26.5-15.73)]

m = 2571/9.661

m = 266 g or 0.266 kg

Hence the mass of the Aluminum fins = 266 g or 0.266 kg

5 0
4 years ago
A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electr
Nata [24]

Answer:

- E = 11.55J

- Q = 0.17C

- E' = (1/4)E

Explanation:

- To calculate the amount of energy stored in the capacitor, you use the following formula:

E=\frac{1}{2}CV^2

C: capacitance = 3800.0*10^-6F

V: potential difference = 78.0V

E=\frac{1}{2}(3800.0*10^{-6}C)(78.0V)^2=11.55J

The energy stored in the capacitor is 11.55J

- If the electrical energy stored in the capacitor is 6.84J, the charge on the capacitor is:

E=\frac{1}{2}QV\\\\Q=\frac{2E}{V}\\\\Q=\frac{2(6.84J)}{78.0V}=0.17C

The charge on the capacitor is 0.17C

- If you take the capacitor as a parallel plate capacitor, you have that the energy stored on the capacitor is:

E=\frac{1}{2}CV^2=\frac{1}{2}(\frac{\epsilon_oA}{d})V^2=\frac{1}{2}\frac{\epsilon_oAV^2}{d}\\\\

A: area of the plates

d: distance between plates

If the distance between plates is increased by a factor of 4, you have:

E'=\frac{1}{2}\frac{\epsilon_oAV^2}{(4d)}=\frac{1}{4}\frac{\epsilon_oAV^2}{2d}=\frac{1}{4}E

Then, the stored energy in the capacitor is decreased by a a factor of (1/4)

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3 years ago
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Answer:

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