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4 years ago

What is the intensity level of a sound with an intensity of 0.000127 W/m2?

1 answer:

6 0

DB = 10 log (I/Io)
Where, Io is the reference intensity which is the minimum intensity that human can hear.
That is; Io = 10^-12
Therefore;
dB = 10 log (0.000127/10^-12) =80.79
= 81 dB
Therefore; the correct answer is 81 dB

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Four identical capacitors are connected with a resistor in two different ways. When they are connected as in part a of the drawi

Nina [5.8K]

Answer:

$T_2 = 0.592$

Explanation:

Given

$T_1 = 1.48s$

See attachment for connection

Required

Determine the time constant in (b)

First, we calculate the total capacitance (C1) in (a):

The upper two connections are connected serially:

So, we have:

$\frac{1}{C_{up}} = \frac{1}{C} + \frac{1}{C}$

Take LCM

$\frac{1}{C_{up}} = \frac{1+1}{C}$

$\frac{1}{C_{up}}= \frac{2}{C}$

Cross Multiply

$C_{up} * 2 = C * 1$

$C_{up} * 2 = C$

Make $C_{up}$ the subject

$C_{up} = \frac{1}{2}C$

The bottom two are also connected serially.

In other words, the upper and the bottom have the same capacitance.

So, the total (C) is:

$C_1 = 2 * C_{up}$

$C_1 = 2 * \frac{1}{2}C$

$C_1 = C$

The total capacitance in (b) is calculated as:

First, we calculate the parallel capacitance (Cp) is:

$C_p = C+C$

$C_p = 2C$

So, the total capacitance (C2) is:

$\frac{1}{C_2} = \frac{1}{C_p} + \frac{1}{C} + \frac{1}{C}$

$\frac{1}{C_2} = \frac{1}{2C} + \frac{1}{C} + \frac{1}{C}$

Take LCM

$\frac{1}{C_2} = \frac{1 + 2 + 2}{2C}$

$\frac{1}{C_2} = \frac{5}{2C}$

Inverse both sides

$C_2 = \frac{2}{5}C$

Both (a) and (b) have the same resistance.

So:

We have:

Time constant is directional proportional to capacitance:

So:

$T\ \alpha\ C$

Convert to equation

$T\ =kC$

Make k the subject

$k = \frac{T}{C}$

$k = \frac{T_1}{C_1} = \frac{T_2}{C_2}$

$\frac{T_1}{C_1} = \frac{T_2}{C_2}$

Make T2 the subject

$T_2 = \frac{T_1 * C_2}{C_1}$

Substitute values for T1, C1 and C2

$T_2 = \frac{1.48 * \frac{2}{5}C}{C}$

$T_2 = \frac{1.48 * \frac{2}{5}}{1}$

$T_2 = \frac{0.592}{1}$

$T_2 = 0.592$

Hence, the time constance of (b) is 0.592 s

8 0

3 years ago

How many electrons can possess this set of quantum numbers: principal quantum number n = 4, magnetic quantum number mℓ = −1?

LekaFEV [45]

Answer:

6 electrons

Explanation:

To solve this question lets determine the possible quantum numbers for the principal quantum number n = 4.

For the quantum number l which describes the shape of the orbital, we have the possible values : 0 to n-1.

Thus, for n = 4 the l can assume the values 0, 1, 2, 3 ( 4 possible shapes )

For the angular quantum number, ml, which tell us the orientation in space,we have the values - l to + l.

So lets determine the number of orbials which can have the values -l for n=4

l = 0 ml = 0

l= 1 ml = -1,0,1

l= 2 ml = -2, -1, 0, 1, 2

l= 3 ml = -3,-2,-1,0,1,2,3

So we have three orbital with ml = - 1 and from Pauli´s exclusion principle we can have up two electrons in each orbital. Thus for n= 4 we can have up to 6 electrons with ml = -1

3 0

4 years ago

Two positive charged particles will ?

Vika [28.1K]

Answer:

make a negative

Explanation:

yep

6 0

3 years ago

Read 2 more answers

A projectile is launched at an angle of 60 degrees to the horizontal at an initial speed of 10 meters per second. What is the ma