A normal conversation involves sound intensities of about 3.0 imes 10⁻⁶ ext Wm⁻². What is the decibel level for this intensity? What is the intensity of the sound for 100dB.

The formula for sound intensity level in decibels ($\beta$) is:

$$\beta = 10 \log_{10} \left(\frac{I}{I_0}\right)$$

where $I$ is the sound intensity and $I_0$ is the reference intensity ($1.0 \times 10^{-12} \text{ Wm}^{-2}$).
Given $I = 3.0 \times 10^{-6} \text{ Wm}^{-2}$.

Substitute the values:

$$\beta = 10 \log_{10} \left(\frac{3.0 \times 10^{-6} \text{ Wm}^{-2}}{1.0 \times 10^{-12} \text{ Wm}^{-2}}\right)$$ $$\beta = 10 \log_{10} (3.0 \times 10^6)$$ $$\beta = 10 (\log_{10} 3.0 + \log_{10} 10^6)$$ $$\beta = 10 (0.4771 + 6)$$ $$\beta = 10 (6.4771)$$ $$\beta = 64.77 \text{ dB}$$

Rearrange the formula to solve for $I$:

$$I = I_0 \times 10^{\beta/10}$$

Given $\beta = 100 \text{ dB}$.

Substitute the values:

$$I = (1.0 \times 10^{-12} \text{ Wm}^{-2}) \times 10^{100/10}$$ $$I = (1.0 \times 10^{-12} \text{ Wm}^{-2}) \times 10^{10}$$ $$I = 1.0 \times 10^{-2} \text{ Wm}^{-2}$$

The decibel level for $3.0 \times 10^{-6} \text{ Wm}^{-2}$ is $\boxed{\text{64.77 dB}}$.
The intensity of sound for 100 dB is $\boxed{1.0 \times 10^{-2} \text{ Wm}^{-2}}$.

Q:2 At 100,000 how much current flows through your body if you touch your terminal of a 12 V battery? If your skin is wet, so that your resistance is only 1000 how much current would you receive from the same battery.

Use Ohm's Law: $I = \frac{V}{R}$
Given: $V = 12 \text{ V}$, $R = 100,000 \ \Omega$.

Substitute the values:

$$I = \frac{12 \text{ V}}{100,000 \ \Omega}$$ $$I = 0.00012 \text{ A}$$ $$I = 0.12 \text{ mA}$$

Given: $V = 12 \text{ V}$, $R = 1000 \ \Omega$.

Substitute the values:

$$I = \frac{12 \text{ V}}{1000 \ \Omega}$$ $$I = 0.012 \text{ A}$$ $$I = 12 \text{ mA}$$

The current for $100,000 \ \Omega$ resistance is $\boxed{0.12 \text{ mA}}$.
The current for $1000 \ \Omega$ resistance is $\boxed{12 \text{ mA}}$.

Q:3 At a particular temperature, the speed of sound in air is $300 \text{ ms}^{-1}$. If the wavelength of a note is 5 cm, calculate the frequency of the sound wave. Is this frequency in the audible range of the human ear?

Given: $\lambda = 5 \text{ cm}$.

$$5 \text{ cm} = 5 \times 10^{-2} \text{ m} = 0.05 \text{ m}$$

The relationship between speed, frequency, and wavelength is $v = f \lambda$.
Rearrange to solve for frequency $f = \frac{v}{\lambda}$.
Given: $v = 300 \text{ ms}^{-1}$, $\lambda = 0.05 \text{ m}$.

Substitute the values:

$$f = \frac{300 \text{ ms}^{-1}}{0.05 \text{ m}}$$ $$f = 6000 \text{ Hz}$$

The audible range for human hearing is typically $20 \text{ Hz}$ to $20,000 \text{ Hz}$.
Since $6000 \text{ Hz}$ falls within this range, it is audible.

The frequency of the sound wave is $\boxed{6000 \text{ Hz}}$.
Yes, this frequency is in the audible range of the human ear.

Q:4 By applying a potential Difference of 1p V across a conductor a current of 1.5 A passes through it. How much energy would be obtain from the current in 2 minutes?

Given: Potential difference $V = 1 \text{ pV}$ (picoVolt).

$$1 \text{ pV} = 1 \times 10^{-12} \text{ V}$$

Given: Time $t = 2 \text{ minutes}$.

$$2 \text{ minutes} = 2 \times 60 \text{ s} = 120 \text{ s}$$

Given: Current $I = 1.5 \text{ A}$.

The formula for electrical energy ($E$) is $E = V I t$.

Substitute the values:

$$E = (1 \times 10^{-12} \text{ V}) \times (1.5 \text{ A}) \times (120 \text{ s})$$ $$E = 1.5 \times 120 \times 10^{-12} \text{ J}$$ $$E = 180 \times 10^{-12} \text{ J}$$ $$E = 1.8 \times 10^{-10} \text{ J}$$

The energy obtained from the current in 2 minutes is $\boxed{1.8 \times 10^{-10} \text{ J}}$.