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JulijaS [17]
2 years ago
12

The diagram shows waves of sound travelling through the air. By which factor is the sound intensity decreased at 3 meters?

Physics
2 answers:
SIZIF [17.4K]2 years ago
8 0
The diagram is missing; however, we know that the intensity of a sound wave is inversely proportional to the square of the distance from the source:
I(r)= \frac{1}{r^2}
where I is the intensity and r is the distance from the source.

We can assume for instance that the initial distance from the source is r=1 m, so that we put 
I= \frac{1}{r^2}= \frac{1}{(1)^2}=1
The intensity at r=3 m will be
I= \frac{1}{r^2}= \frac{1}{(3)^2}= \frac{1}{9}
Therefore, the sound intensity has decreased by a factor 1/9.
RSB [31]2 years ago
3 0

its a  on edge just finished my exam

heres the diagram

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katen-ka-za [31]

The initial potential energy of the wagon containing gold boxes will enable

it roll down the hill when cut loose.

The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.

Reasons:

Mass wagon and gold = 166 kg

Location of the wagon = 77 meters up the hill

Slope of the hill = 8°

Location of the rangers = 41 meters from the canyon

Mass of Lone Ranger, m₁ = 65 kg

Mass of Tonto m₂ = 66 kg

Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}

Where;

v = The velocity of the wagon a the bottom of the cliff

Therefore;

\dfrac{1}{2} \times 166 \times v^2 = 17457.0912

v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5

Velocity of the wagon, v ≈ 14.5 m/s

Momentum = Mass, m × Velocity, v

Initial momentum of wagon = m·v

Final momentum of wagon and ranger = (m + m₁ + m₂)·v'

By conservation of momentum, we have;

m·v = (m + m₁ + m₂)·v'

\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2)  }}

Which gives;

\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66)  } \approx 8.1

The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s

Time = \dfrac{Distance}{Velocity}

\mathrm{The \ time \ the\ Lone \  Ranger \  and  \ Tonto \  have,  \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s

The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

Learn more here:

brainly.com/question/11888124

brainly.com/question/16492221

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2 years ago
NEED HELP ASAP
postnew [5]

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The gravity of Neptune is about 1.1 times the gravity of earth. How will the mass of an object on Neptune compare with its mass
Roman55 [17]
I think the gravity doesn't affect the mass of an object. Only it's weight can be compared
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2 years ago
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A grandfather clock is "losing" time because its pendulum moves too slowly. Assume that the pendulum is a massive bob at the end
IgorC [24]

Answer:

d) shortening the string

Explanation:

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When a clock loses time, the time period of the pendulum clock increases.

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3 years ago
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An object weighing 49 N is dropped from a height of 30 m. It is found to be moving with a velocity of 24m/s just before it hits
faltersainse [42]

The  frictional force will be 0.22N.

<h3>What is Frictional force?</h3>

Frictional force is the force generated between two surfaces that are in contact and slide against each other.

Given,

Weight=4N

mass =4.9/9.8=0.5kg

Hieght =30m

velocity=24m/s

Acceration , v²-u²=2as

24²/2×30 =a , u is zero

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By Using conservation  of energy ,

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30F=1150-144

F= 6/30

F=0.2N

The force will be 0.2N

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