15. An e.mf. of 16 volts is induced in a coil of inductance 4H. The rate of change of current must be:

To solve this problem we need to apply the corresponding sound intensity measured from the logarithmic scale. Since in the range of intensities that the human ear can detect without pain there are large differences in the number of figures used on a linear scale, it is usual to use a logarithmic scale. The unit most used in the logarithmic scale is the decibel yes described as

$\beta_{dB} = 10log_{10} \frac{I}{I_0}$

Where,

I = Acoustic intensity in linear scale

$I_0$ = Hearing threshold

The value in decibels is 17dB, then

$17dB = 10log_{10} \frac{I}{I_0}$

Using properties of logarithms we have,

$\frac{17}{10} = log_{10} \frac{I}{I_0}$

$log_{10} \frac{I}{I_0} = 1.7$

$\frac{I}{I_0} = 10^{1.7}$

$\frac{I}{I_0} = 50.12 W/m^2$

Therefore the factor that the intensity of the sound was $50.12W/m^2$

5 0

3 years ago

The electric field strength in the space between two closely spaced parallel disks is 1.0 105 N/C. This field is the result of t

balu736 [363]

Answer:

$D=2.996\times 10^{-2} m$

Explanation:

*Assume the parallel disks have equal diameters.

Given the electric strength as $1.0\times 10^5 N/C.$ transferring $3.9\times 10^9$ electrons, the disk's Area can be calculated using the formula:

$E=\frac{\eta}{\epsilon_o}=\frac{Q}{A\epsilon_o}\\\\A=\frac{Q}{E\epsilon_o}\\\\=\frac{(3.9\times 10^9)\times (1.6\times10^{-19})}{(1.0\times 10^5 )\times (8.85\times10^{-12})}\\\\A=7.0508\times 10^{-4} \ m^2$

#We now calculate the disks diameter:

$A=\pi(D/2)^2\\\\2\sqrt{\frac{A}{\pi}}=D\\\\=2\sqrt{7.0508\times 10^{-4}/\pi}\\\\D=2.996\times 10^{-2} \ m$

Hence, the diameter of the disks is $D=2.996\times 10^{-2} m$

8 0

4 years ago

A wire is joined to points X and Y in the circuit diagram shown. A diagram of a circuit with a power source on the left. Directl

Sidana [21]

The circuit change when the wire is added will see a short circuit occur and makes bulbs 1 and 2 turn off but keeps bulbs 3 and 4 lit. Option D. This is further explained below.

<h3>How does the circuit change when the wire is added?</h3>

Generally, Electronic circuits consist of a series of interconnected parts that form a closed loop through which electricity may flow.

In conclusion, If two wires are linked together, a short circuit will develop, cutting power to bulbs 1 and 2. But there is no impact on bulbs 3 and 4. There is no problem with bulbs 3 and 4.