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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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Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal ener
AVprozaik [17]

Answer:

1838216 J

Explanation:

95 km/h = 26.39 m/s

40 km/h = 11.11 m/s

Initial kinetic energy

= .5 x 1600 x(26.39)²

= 557145.67 J

Final kinetic energy

= .5 x 1600 x ( 11.11)²

= 98745.68 J

Loss of kinetic energy

= 458400 J

Loss of potential energy

= mg x loss of height

= 1600 x 9.8 x 340 sin 15

= 1379816 J

Sum of Loss of potential energy and Loss of kinetic energy

=  1379816 + 458400

= 1838216 J

This is the work done by the friction . So this is heat generated.

8 0
3 years ago
3. A sprinter leaves the starting blocks with an acceleration of 4.5 m/s2. What is the
UkoKoshka [18]

Hi there! :)

\large\boxed{v_{f} = 18 m/s}

Use the following kinematic equation to solve for the final velocity:

v_{f} = v_{i} + at

In this instance, the runner started from rest, so the initial velocity is 0 m/s. We can rewrite the equation as:

v_{f} = at

Plug in the given acceleration and time:

v_{f} = 4.5 * 4 = 18 m/s

5 0
3 years ago
1. What is the gravitational force of attraction between two 800 kg spherical objects that are 5 m apart? What would the gravita
Katyanochek1 [597]

Answer:

the force of gravity between them is quadrupled .

Explanation:

Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces.

4 0
3 years ago
Electro-magnets are created by...
mojhsa [17]
Wrapping in insulated wire around a medal with ferromagnetic properties and applying an electric current
6 0
3 years ago
A particular heat engine has a mechanical power output of 4.00 kW and an efficiency of 26.0%. The engine expels 8.55 103 J of ex
Ivahew [28]

To develop the problem we will start by finding the energy taken by each cycle through the efficiency of the motor and the exhausted energy. Later the work will be found for the conservation of energy in which this is equivalent to the difference between the two calculated energy values. Finally the estimated time will be calculated with the work and the power given,

\text{Efficiency of the heat engine} = \eta = 26\% = 0.26

\text{Energy taken in by the heat engine during each cycle} = Q_h

\text{Energy exhausted by the heat engine in each cycle} = Q_c = 8.55*10^3 J

\eta = 1 - \frac{Q_{c}}{Q_{h}}

0.26 = 1 - \frac{8.55\ast 10^{3}}{Q_{h}}

\frac{8.55* 10^{3}}{Q_{h}} = 0.74

Q_h = \frac{8.55*10^3}{0.74}

Q_h = 11.554*10^3J

PART A)

Work done by the heat engine in each cycle = W

W = Q_h-Q_c

W = 11.554*10^3J-8.55*10^3J

W = 3004J

According to the value given we have that,

P = 4.0kW

P = 4000W

Power is defined as the variation of energy as a function of time therefore,

P = \frac{W}{t}

4000W = \frac{3004J}{t}

t = \frac{3004}{4000}

t = 0.75s

Therefore the interval for each cycle is 0.75s

5 0
3 years ago
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