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krek1111 [17]
3 years ago
9

You are listening to the radio when one of your favorite songs comes on, so you turn up the volume. If you managed to increase t

he sound intensity by
17.0 dB, by what factor did the intensity of the sound, in W/m2, increase?
The intensity of sound, in W/m2, increased by a factor of .
Physics
1 answer:
andrew-mc [135]3 years ago
5 0

To solve this problem we need to apply the corresponding sound intensity measured from the logarithmic scale. Since in the range of intensities that the human ear can detect without pain there are large differences in the number of figures used on a linear scale, it is usual to use a logarithmic scale. The unit most used in the logarithmic scale is the decibel yes described as

\beta_{dB} = 10log_{10} \frac{I}{I_0}

Where,

I = Acoustic intensity in linear scale

I_0 = Hearing threshold

The value in decibels is 17dB, then

17dB = 10log_{10} \frac{I}{I_0}

Using properties of logarithms we have,

\frac{17}{10} = log_{10} \frac{I}{I_0}

log_{10} \frac{I}{I_0} = 1.7

\frac{I}{I_0} = 10^{1.7}

\frac{I}{I_0} = 50.12 W/m^2

Therefore the factor that the intensity of the sound was 50.12W/m^2

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During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 6.22 s, how high does it rise? The accel
BigorU [14]

Answer:

47.4 m

Explanation:

When an object is thrown upward, it rises up, it reaches its maximum height, and then it goes down. The time at which it reaches its maximum height is half the total time of flight.

In this case, the time of flight is 6.22 s, so the time the ball takes to reach the maximum height is

t=\frac{6.22}{2}=3.11 s

Now we consider only the downward motion of the ball: it is a free fall motion, so we can find the vertical displacement by using the suvat equation

s=ut+\frac{1}{2}gt^2

where

s is the vertical displacement

u = 0 is the initial velocity

t = 3.11 s is the time

g=9.8 m/s^2 is the acceleration of gravity (taking downward as positive direction)

Solving the  formula, we find

s=\frac{1}{2}(9.8)(3.11)^2=47.4 m

7 0
3 years ago
What is the difference between a measurement and an observation?A. Measurements are made with tools; observations are not.B. Obs
choli [55]

Answer:

A. Measurements are made with tools; observations are not

Explanation:

Measurement: <em>the assignment of numbers or codes according to prior-set rules. </em>

Observation: <em>data from an individual study subject or sampled unit.</em>

Measurement error: <em>differences between "true" answers and what appears on data collection instruments</em>

hope this helps

plz mark brainleist

8 0
3 years ago
The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou
Mama L [17]

A) Time needed: 6.24 s

B) Time needed: 2.86 s

Explanation:

A)

In this part, we are told that the power if the engine is constant. The power of the engine is given by

P=\frac{W}{t}

where

W is the work done

t is the time

This means that the power of the engine is proportional to the work done, and therefore, to the kinetic energy of the car:

P=\frac{\frac{1}{2}mv^2}{t}=const.

where m is the mass of the car and v its velocity.

SInce power is constant, we can write:

\frac{\frac{1}{2}mv_1^2}{t_1^2}=\frac{\frac{1}{2}mv_2^2}{t_2}

where:

t_1=1.40 s is the time the car needs to accelerates to v_1=28.0 mph

t_2 is the time the car needs to accelerate to v_2=57.0 mph

Therefore, solving for t_2,

t_2 = \frac{v^2}{u^2}t_1=\frac{57^2}{28^2}(1.40)=6.24 s

B)

First of all, we have to calculate the acceleration of the car. We can do it using the following equation:

a=\frac{v-u}{t}

where:

u = 0 is the initial velocity

v=28.0 mph \cdot \frac{1609 m/mi}{3600 s/h}=12.5 m/s is the final velocity

t = 1.40 s is the time elapsed

Substituting, we find the acceleration:

a=\frac{12.5-0}{1.40}=8.9 m/s^2

In this part, we are told that the force exerted by the engine is constant: according to Newton's second law, acceleration is proportional to the force,

F=ma

This means that the acceleration is also constant.

Now we want to find how long the car takes to accelerate to a final velocity of

v=57.0 mph \cdot \frac{1609}{3600}=25.5 m/s

From an initial velocity of

u = 0

Using again the same suvat equation, and using the acceleration we found previously, we find:

t=\frac{v-u}{a}=\frac{25.5-0}{8.9}=2.87 s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

About power:

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#LearnwithBrainly

6 0
3 years ago
Which is most likely the length of a student’s textbook?
UkoKoshka [18]

For this case, what we must do is to rewrite these measurements in the same unit in order to compare them.

By writing the measurements in meters we have:

30 mm = (30) * (\frac{1}{1000}) = 0.030 m

30 cm = (30) * (\frac{1}{100}) = 0.30 m

30 dm = (30) * (\frac{1}{10}) = 3 m\\30 Hm = (30) * (100) = 3000 m

Therefore, physically the correct measure is:

0.30 m = 30 cm

Answer:

the length of a student's textbook most likely is:

30 centimeters

3 0
3 years ago
Read 2 more answers
*WILL MARK BRAINLIEST FOR RIGHT ANSWER* How much current must be applied across a 60 Ω light bulb filament in order for it to co
sp2606 [1]

Answer: The current must be equal to \frac{\sqrt{33} }{6} amps, or ~0.9574 amps.

Explanation:

You can find the current in amperes using ohms and watts from this formula:

I = \sqrt{\frac{P}{R} }

Where P represents power in watts, R represents resistance in ohms, and I represents current in amperes.

You can then substitute 60 and 55 into the equation to find I:

I = \sqrt{\frac{55}{60} } \\I = \frac{\sqrt{55} }{\sqrt{60} }

Then, simplify the denominator:

I = \frac{\sqrt{55} }{2\sqrt{15} }

Rationalize the denominator:

I = \frac{\sqrt{55} }{2\sqrt{15} } * \frac{\sqrt{15} }{\sqrt{15} } = \frac{\sqrt{825} }{30}

Simplify the numerator by finding its factors:

I = \frac{5\sqrt{33} }{30} = \frac{\sqrt{33} }{6}

The current must be equal to \frac{\sqrt{33} }{6} amps, or ~0.9574 amps.

8 0
3 years ago
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