All categories

3 years ago

Sound level B in decibels is defined as

Physics

1 answer:

Lelechka [254]3 years ago

3 0

Answer:

The approximate combined sound intensity is $I_{T}=1.1\times10^{-4}W/m^{2}$

Explanation:

The decibel scale intensity for busy traffic is 80 dB. so intensity will be

$10log(\frac{I_{1}}{I_{0}} )=80$ , therefore $I_{1}=1\times10^{8}I_{0}=1\times10^{8} * 1\times10^{-12}W/m^{2}=1\times10^{-4}W/m^{2}$

In the same way for the loud conversation having a decibel intensity of 70 dB.

$10log(\frac{I_{2}}{I_{0}} )=70$ , therefore $I_{2}=1\times10^{7}I_{0}=1\times10^{7} * 1\times10^{-12}W/m^{2}=1\times10^{-5}W/m^{2}$

Finally we add both of them $I_{T}=I_{1}+I_{2}=1\times10^{-4}W/m^{2}+1\times10^{-5}W/m^{2}=1.1\times10^{-4}W/m^{2}$ , is the approximate combined sound intensity.

You might be interested in

GIVING BRAINLIEST FIVE STARS AND HEART!

alina1380 [7]

Answer:

A bicycle on the top of the hill has the highest potential energy, and when the bike goes down, it transfers to kinetic because it is moving

Explanation:

yeah

4 0

3 years ago

The inner conductor of a coaxial cable has a radius of 0.800 mm, and the outer conductor’s inside radius is 3.00 mm. The space b

ZanzabumX [31]

Answer:

The maximum potential difference is 186.02 x 10¹⁵ V

Explanation:

formula for calculating maximum potential difference

$V = \frac{2K_e \lambda}{k}ln(\frac{b}{a})$

where;

Ke is coulomb's constant = 8.99 x 10⁹ Nm²/c²

k is the dielectric constant = 2.3

b is the outer radius of the conductor = 3 mm

a is the inner radius of the conductor = 0.8 mm

λ is the linear charge density = 18 x 10⁶ V/m

Substitute in these values in the above equation;

$V = \frac{2K_e \lambda}{k}ln(\frac{b}{a}) = \frac{2*8.99*10^9*18*10^6 }{2.3}ln(\frac{3}{0.8}) =140.71 *10^{15} *1.322 \\\\V= 186.02 *10^{15} \ V$

Therefore, the maximum potential difference this cable can withstand is 186.02 x 10¹⁵ V

8 0

3 years ago

(I) In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.6 cm. A second pro

Kipish [7]

Answer:

The second projectile was 1.41 times faster than the first.

Explanation:

In the ballistic pendulum experiment, the speed (v) of the projectile is given by:

$v = \frac{m + M}{m} \cdot \sqrt{2gh}$

<em>where m: is the mass of the projectile, M: is the mass of the pendulum, g: is the gravitational constant and h: is the maximum height of the pendulum. </em>

To know how many times faster was the second projectile than the first, we need to take the ratio for the velocities for the projectiles 2 and 1:

$\frac{v_{2}}{v_{1}} = \frac{\frac{m_{2} + M}{m_{2}} \cdot \sqrt{2gh_{2}}}{\frac{m_{1} + M}{m_{1}} \cdot \sqrt{2gh_{1}}}$ (1)

<em>where m₁ and m₂ are the masses of the projectiles 1 and 2, respectively, and h₁ and h₂ are the maximum height reached by the pendulum by the projectiles 1 and 2, respectively. </em>

Since the projectile 1 has the same mass that the projectile 2, we can simplify equation (1):

$\frac{v_{2}}{v_{1}} = \frac{\sqrt{h_{2}}}{\sqrt{h_{1}}}$

$\frac{v_{2}}{v_{1}} = \frac{\sqrt{5.2 cm}}{\sqrt{2.6 cm}}$

$\frac{v_{2}}{v_{1}} = 1.41$

Therefore, the second projectile was 1.41 times faster than the first.

I hope it helps you!

8 0

3 years ago

What is the surface area to volume ratio of this cube

tamaranim1 [39]

I can't see that cube from here.

But if the length of the side of the cube is ' K ' units,
then the surface area of the cube is 6K² units², and
the volume of the cube is K³ units³.

The ratio of the surface area to the volume is

(6K² units²) / (K³ units³) = (6) / (K units) .

So for example, if the side of the cube is 2 inches, then
the ratio of surface area to volume is "3 per inch".

That's the answer. I did the whole thing in order to earn
the points, but I don't expect you to understand much of it,
because I see from your username that you suck at math.
I'm sorry you decided that. Now that you've put up the
brick wall, it'll be even harder for any math to find its way
in there, and you'll miss out on a lot of the fun.

3 0

3 years ago

Explain the differences between horticulture and agriculture.

sammy [17]

<h2><u>Hello </u><u>There</u></h2>

$\tt{Difference \: Between \: \underline\red {Horticulture} \: And \: \underline\green{Agriculture}}$

$\text{\underline\red {Horticulture}}$

The Cultivation of Fruits, Vegetables and Flowers for domestic and international markets are called Horticulture.

$\text{\underline\green{Agriculture}}$

It is science, art and occupation of cultivating the soil, producing crops and livestock. It is systematic and controlled use of living organisms and the environment to improve the human condition.

<h3>Hope This Helps</h3>

6 0

2 years ago