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grin007 [14]
3 years ago
5

a person when asked to speak up,increases her sound level from 30dB to 60dB.The amount of power per unit area increased by? a)30

00 times b)30 times c)2 times d)1000 times
Physics
1 answer:
aivan3 [116]3 years ago
4 0

Answer:

d) 1000 times

Explanation:

As we know that difference of sound level is given as

L_2 - L_1 = 10 Log \frac{I_2}{I_1}

so here we need to find the ratio of two intensity

it is given as

Log\frac{I_2}{I_1} = \frac{(L_2- L_1)}{10}

Log\frac{I_2}{I_1} = \frac{60 - 30}{10}

Log\frac{I_2}{I_1} = 3

now we have

\frac{I_2}{I_1} = 10^3

so it is

d) 1000 times

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The sound intensity at a distance 2.00 m from a sound source is 5.00 Find the total sound energy emitted by the source in each s
notka56 [123]

Answer:

     P = 251, 3 W

Explanation:

The intensity is defined as the power emitted per unit area

           I = P / A

Since sound is distributed in all directions spherical shape, the area of ​​a sphere is

           A = 4π r²

let's clear the power and replace

         P = I A

         P = I (4π r²)

let's calculate

         P = 5.00 (4π 2²)

         P = 251, 3 W

6 0
3 years ago
A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o
inysia [295]

Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) =  \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2  \omega_i^2) =  \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2  \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2  ) =  \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 -  \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 -  \frac{3}{4}v_f^2\\\\

h = \frac{3}{4g}(v_1^2 -v_f^2)

given;

v₁ = 5.0 m/s

vf = 0

g = 9.8 m/s²

h = \frac{3}{4g}(v_1^2 -v_f^2) =\frac{3}{4*9.8}(5^2 -0) = 1.91 \ m

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m

5 0
3 years ago
A tin can collapses if all air inside it is taken out why
Veseljchak [2.6K]

That only happens when the tin can is IN air.

In the familiar, comfy part of Earth's atmosphere where we live, the normal pressure of air is around 14.6 pounds on every square inch of everything. That's a big part of the reason why we're built with bodies that generate that same amount of pressure on the INSIDE pressing OUT. That way, we always have the same pressure pushing in both directions, so we know that we won't get crushed or blow up like balloons.

But we have to be careful with our bodies or other things when they're in places where the atmospheric pressure on the outside is NOT normal.

-- When a deep-sea diver goes hundreds of feet down in the ocean, and the pressure of the water is much GREATER than normal air.

-- When an astronaut has to go outside ... where there's NO air ... and fix something on the International Space Station.

When the pressure on the outside becomes very unusual, we have to wear special suits to protect our bodies from the unusual conditions.

The tin can in the story is a lot like our bodies. As long as it has air inside and air outside, the pressure is the same in both directions, so there's no particular force trying to deform the can. But ...

-- If you seal the can with the air inside it, take the can into a vacuum chamber, and pump the air out of the vacuum chamber, then the can only has pressure inside. It'll expand, and eventually spring a little hole in the metal, and all the air inside will blow out.

-- If you take all the air OUT of the can (so the can is REALLY 'empty'), then the pressure on it is all from the outside. In that situation, the can simply collapses, because there's nothing inside to provide pressure in the outward direction.

One more little thing to think about:

When you want some toothpaste to come drizzling out of the tube onto your brush, what do you do ? Do you perhaps squeeze the tube, and increase the pressure on the outside ?

4 0
3 years ago
The total negative charge on the electrons in 1mol of helium (atomic number 2, molar mass 4) is ________?
DaniilM [7]

Answer:

1.92\times 10^5 C

Explanation:

We are given that

Atomic number=2

We have to find the total negative charge on the electrons in one mole of Helium.

We know that atomic number=Proton number

Proton number=Number of electrons=2

Number of electrons in Helium=2

1 mole of Helium=6.02\times 10^{23} atoms

We know that q=ne

Where n =Number of fundamental units

e=Charge on electron

1 e=1.6\times 10^{-19}C

Using the formula

q=2\times 1.6\times 10^{-19}=3.2\times 10^{-19}C

Total negative charge in 1 mole=3.2\times 10^{-19}\times 6.02\times 10^{23}=1.92\times 10^5C

Hence, the total negative charge on the electrons in 1 mole of Helium=1.92\times 10^5 C

4 0
3 years ago
What property is an acid-base indicator used to measure?
Volgvan

Answer:

ph

Explanation:

4 0
3 years ago
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