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KIM [24]
3 years ago
8

The sound intensity at the ear of passenger in a car with a damaged muffler is 8.0 × 10-3 W/m2. What is the intensity level of t

his sound in decibels? Use the threshold of hearing (1.0 × 10-12 W/m2) as the refere
Physics
1 answer:
Mama L [17]3 years ago
4 0

Answer:

99 dB

Explanation:

We have given that sound intensity with a damaged muffler =8× 10^{-3}

we have to find the intensity level of the this sound in decibels

for calculating in decibels we have to use the formula

β=10log\frac{I}{I_0}

  =10log\frac{.008}{10^{-12}}

   =10 log8+10 log10^{9}

   =10 log8+90 log10

   =10×0.9030+90

   =99 dB

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An object at rest starts accelerating.
Shalnov [3]

Answer:

<u>We are given: </u>

initial velocity (u) = 0 m/s

final velocity (v) = 10 m/s

displacement (s) = 20 m

acceleration (a) = a m/s/s

<u>Solving for 'a'</u>

From the third equation of motion:

v² - u² = 2as

replacing the variables

(10)² - (0)² = 2(a)(20)

100 = 40a

a = 100 / 40

a = 2.5 m/s²

6 0
3 years ago
A uniform disk, a thin hoop, and a uniform sphere, all with the same mass and same outer radius, are each free to rotate about a
olga nikolaevna [1]

Answer:

4 hoop, disk, sphere

Explanation:

Because

We are given data that

Hoop, disk, sphere have Same mass and radius

So let

And Initial angular velocity, = 0

The Force on each be F

And Time = t

Also let

Radius of each = r

So let's find the inertia shall we!!

I1 = m r² /2

= 0.5 mr² the his is for dis

I2 = m r² for hoop

And

Moment of inertia of sphere wiil be

I3 = (2/5) mr²

= 0.4 mr²

So

ωf = ωi + α t

= 0 + ( τ / I ) t

= ( F r / I ) t

So we can see that

ωf is inversely proportional to moment of inertia.

And so we take the

Order of I ( least to greatest ) :

I3 (sphere) , I1 (disk) , I2 (hoop) , ,

Order of ωf: ( least to greatest)

That of omega xf is the reverse of inertial so

hoop, disk, sphere

Option - 4

5 0
4 years ago
8. Three grams of Bismuth-218 decay to 0.375 grams in one hour. What is the half-
Evgen [1.6K]

Answer: 0.333 h

Explanation:

This problem can be solved using the <u>Radioactive Half Life Formula</u>:  

A=A_{o}.2^{\frac{-t}{H}} (1)  

Where:  

A=0.375 g is the final amount of the material  

A_{o}=3 g is the initial amount of the material  

t=1 h is the time elapsed  

H is the half life of the material (the quantity we are asked to find)  

Knowing this, let's substitute the values and find h from (1):

0.375 g=(3 g)2^{\frac{-1h}{H}} (2)  

\frac{0.375 g}{3 g}=2^{\frac{-1h}{H}} (3)  

Applying natural logarithm in both sides:

ln(\frac{0.375 g}{3 g})=ln(2^{\frac{-1 h}{H}}) (4)  

-2.079=-\frac{1 h}{H}ln(2) (5)  

Clearing H:

H=\frac{-1h}{-2.079}(0.693) (6)  

Finally:

h=0.333 h This is the half-life of the Bismuth-218 isotope

4 0
4 years ago
11.
jolli1 [7]
<h3>Reducing Surface Area.</h3>

If the surface area becomes smaller, the pressure becomes larger.

8 0
3 years ago
Please help ASAP.
Bess [88]

Im not 100% sure you have to tell me if im wrong or not.

D

B

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3 0
3 years ago
Read 2 more answers
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