Can someone do this for the football

What is the intensity of a sound with a sound intensity level (SIL) 67 dB, in units of W/m^2?

vfiekz [6]

Answer:

The intensity of sound (I) = 3.16 x 10⁻⁶ W/m²

Explanation:

We have expression for sound intensity level (SIL),

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

Here we need to find the intensity of sound (I).

$L=10log_{10}\left ( \frac{I}{I_0}\right )\\\\log_{10}\left ( \frac{I}{I_0}\right )=0.1L\\\\\frac{I}{I_0}=10^{0.1L}\\\\I=I_010^{0.1L}$

Substituting

L = 67 dB and I₀ = 10⁻¹² W/m² in the equation

$I=I_010^{0.1L}=10^{-12}\times 10^{0.1\times 65}\\\\I_0=10^{-12}\times 10^{6.5}=10^{-5.5}=3.16\times 10^{-6}W/m^2$

The intensity of sound (I) = 3.16 x 10⁻⁶ W/m²

8 0

3 years ago

What kind of friction exists between solid objects moving in water?

OLga [1]

In water, it's drag. Exactly the same as air resistance
only hundreds of times greater.

5 0

3 years ago

Water has a mass per mole of 18.0 g/mol, and each water mo- lecule (H2O) has 10 electrons. (a) How many electrons are there in o

Dovator [93]

Answer:

a) There are $3,35*10^{26}$ electrons in a liter of water.

b) The net charge is -53601707,1 C

Explanation:

a) To find out how many electrons are in a liter of water (equivalent to 1000 grams of water), we have to find out how many molecules of water there are and then multiply it by 10 (e- per molecule).

We can find out how many molecules are by finding the number of moles and then multiplying it by Avogadro's number (number of elements per mol):

$e^{-}= \frac{m_{water} }{M_{water} } * Avogadro* \frac{e^{-} }{molecule}= \frac{1000g }{18g/mol} * 6,022*10^{23} * \frac{10e^{-} }{molecule}=3,35*10^{26} e^{-}$

b) As all electrons have the same charge, in order to find the net charge of those electrons we have to multiply the charge of a single electron by the number of electrons:

$Net charge= -1,602*10^{-19} \frac{C}{e^{-} } *3,35*10^{26} e^{-}=-53601707,1 C$

An important clarification is that while the net charge may seem huge, water as a whole is a neutral medium, because there are as many protons as there are electrons, and as they have the same charge, the net charge of water is 0.

8 0

4 years ago

In the nucleus, the a)...... have b)...... charge.

Inga [223]

C:carry
I hope I help

8 0

4 years ago

A car travels along a straight road, heading east for 1 h, then traveling for 30 min on another road that leads northeast. If th

Bogdan [553]

Answer:

The car is 72.75 miles away from its starting position.

Explanation:

First, remember the relation:

distance = time*speed.

Also, the distance between two points (a, b) and (c, d) is:

D = √( (a - c)^2 + (b - d)^2)

For this problem, we can assume:

The North is equivalent to the y-axis, and the East is equivalent to the x-axis.

We also assume that the initial position of the car is (0mi, 0mi)

Now the car moves to the East at a speed of 52mi/h for one hour, then the new position of the car is:

(0mi, 0mi) + (52mi/h*1h, 0mi) = (52mi, 0mi)

Now the car travels 30 mins (or 0.5 hours) to the northeast at a speed of 52mi/h.

We can assume that it moves at an exact angle of 45° from East to North, then the components of the speed can be written as:

Sx = speed in the x-axis = 52mi/h*cos(45°) = 36.77 mi/h

Sy = speed in the y-axis = 52mi/h*sin(45°) = 36.77 mi/h

Then the new position of the car is:

(52mi, 0mi) + (36.77 mi/h*0.5h, 36.77 mi/h*0.5h) = (70.385 mi, 18.385 mi)

Now we know the final position of the car.

The distance between the final position (70.385 mi, 18.385 mi) and the initial position (0mi, 0mi) is:

D = √( (70.385 mi - 0mi)^2 + (18.385 mi - 0mi)^2) = 72.75 mi

The car is 72.75 miles away from its starting position.

7 0

3 years ago