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3 years ago

The kind of graph that would best show the

Physics

1 answer:

Semenov [28]3 years ago

6 0

Answer:

a bar graph would be the kind of graph that you should use for this problem because it would be easier to read and also do

Explanation

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The intensity of the sound from a certain source is measured at two points along a line from the source. The points are separate

Artemon [7]

Answer:

The source is at a distance of 4.56 m from the first point.

Solution:

As per the question:

Separation distance between the points, d = 11.0 m

Sound level at the first point, L = 66.40 dB

Sound level at the second point, L'= 55.74 dB

Now,

$L = 10log_{10}\frac{I}{I_{o}}$

$I = I_{o}10^{\frac{L}{10}} = I_{o}10^{0.1L} = 10^{- 12}\times 10^{0.1\times 66.40} = 10^{- 5.36}$

$L' = 10log_{10}\frac{I'}{I_{o}}$

$I' = I_{o}10^{\frac{L'}{10}} = 10^{- 12}\times 10^{0.1\times 55.74} = 10^{- 6.426}$

where

$I_{o} = 10^{- 12} W/m^{2}$

I = Intensity of sound

Now,

$I = \frac{P}{4\pi R^{2}}$

Similarly,

$I' = \frac{P}{4\pi (R + 11.0)^{2}}$

Now,

$\frac{I}{I'} = \frac{(R + 11.0)^{2}}{R^{2}}$

$\frac{10^{- 5.36}}{10^{- 6.426}} = \frac{(R + 11.0)^{2}}{R^{2}}$

$R ^{2} + 22R + 121 = 11.64R^{2}}$

$10.64R ^{2} - 22R - 121 = 0$

Solving the above quadratic eqn, we get:

R = 4.56 m

8 0

3 years ago

Since the noble gases are stable (VERY unreactive), oxygen would like to look like one in terms of the number and arrangement of

iogann1982 [59]

Answer:

oxygen gains two electron to attain electron configuration like neon. Oe-²

Explanation:

3 0

3 years ago

Vector sum of 15 m, East and 9 m, West is 24 m, East. True False

Firlakuza [10]

Answer:

False

Explanation:

Because when you go through east

( +x axis ) then you go to west ( -x axis )

You will subtract -9 from +15

it's become +6

( I talk about the displacement not distance) ( West = - East )

I hope that it's a clear ") .

7 0

3 years ago

Read 2 more answers

During a race, a runner runs at a speed of 6 m/s. Four seconds later, she is running at a speed of 14 m/s. What is the runner's

DedPeter [7]

Answer:

D. 2 m/s²

Step-by-step explanation:

Initial speed of the runner (u) = 6 m/s

Final speed of the runner (v) = 14 m/s

Time taken (t) = 4 s

By using equation of motion, we get:

$\bf \longrightarrow v = u + at \\ \\ \rm \longrightarrow 14 = 6 + 4a \\ \\ \rm \longrightarrow 14 - 6 = 6 - 6 + 4a \\ \\ \rm \longrightarrow 8 = 4a \\ \\ \rm \longrightarrow 4a = 8 \\ \\ \rm \longrightarrow \dfrac{4a}{4} = \dfrac{8}{4} \\ \\ \rm \longrightarrow a = 2 \: m {s}^{ - 2}$

$\therefore$ Acceleration of the runner (a) = 2 m/s²

3 0

3 years ago

In a Millikan oil-drop experiment, the condenser plates are spaced 2.00 cm apart, the potential across the plates is 4000 V, the

astraxan [27]

Answer:

Answer is in the following attachment

Explanation:

3 0

3 years ago