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3 days ago

If one firecracker produces a sound intensity of 90db, what would be the intensity of a sound produced by 1000 firecrackers, all

going off at once?

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Maru [3345]

Answer:

$\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K$

Consequently, the temperature difference across the material will be $\Delta T = 375 K$

Explanation:

In this case, we apply the Fourier Law of heat conduction expressed by the following equation:

$Q = -kA \frac{\Delta T}{\Delta x}$ (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 denotes the cross-sectional area

Q= 3KW signifies the heat transfer rate

$\Delta T$ is the temperature difference we need to determine

represents the thickness of the material $\Delta x=2.5 cm =0.025 m$

To isolate $\Delta T$ from equation (1), we obtain:

$\Delta T =\frac{Q \Delta x}{Ak}$

Initially, we convert 3KW to W, resulting in:

$Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W$

With all variables accounted for, we can substitute and calculate:

$\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K$

Thus, the temperature difference across the material will be $\Delta T = 375 K$

5 0

2 months ago

Softa [3030]

The sheet's charge density calculates to be 1.384×10⁻⁷C/m².

Charge density is the ratio of charge per unit area.

The square sheet has dimensions l=17 cm.

To compute the area A of the sheet.

$A=l^2=(17 cm)^2= (17*10^-^2m)^2=0.0289 m^2$

The overall charge Q on the sheet is

$Q=4nC=4*10^-^9C$

The charge density σ is defined as

$\sigma=\frac{Q}{A}$

We substitute 4×10⁻⁹C for Q and 0.0289 m² for A.

$\sigma=\frac{Q}{A}\\ =\frac{4*10^-^9C}{0.0289 m^2} \\ =1.389*10^-^7C/m^2$

Thus, the charge density for the sheet is 1.384×10⁻⁷C/m².

8 0

2 months ago