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tresset_1 [31]
3 years ago
7

A string of length 0.6 M is vibrating at 100 Hz and its second harmonic and producing sound that moves at 340 m/s. What is true

about the frequent or wavelength of this sound?
A) The wavelength of the sound is 0.6 m.

B) The frequency of the sound is 100 Hz.

C) The frequency of the sound is 567 Hz.

D) The wavelength of the sound is 0.9 m.
Physics
2 answers:
ivanzaharov [21]3 years ago
7 0

Answer:

B) The frequency of the sound is 100 Hz.

Explanation:

Frequency of the sound is always equal to the frequency of the source

So here we know that

length of the string is

L =0.6 m

Speed of the sound is

v = 340 m/s

frequency of the sound

f = 100 Hz

now the wavelength of the sound is given as

\lambda = \frac{v}{f}

\lambda = \frac{340}{100}

\lambda = 3.40 m

Cloud [144]3 years ago
6 0

Answer: the answer is B

Explanation: I just took the quiz :)

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Identify two potential improvements to the opal extraction process and explain how these improvements could minimize harm to the
Orlov [11]

Answer

• Improving the environmental performances

• Developing Green Mining technology

Explanation

The effect to the environment caused by opal mining are; impact on soils and geology, clearing of native vegetation disrupting flora and fauna, change in land use and effects of air quality.

Opal mining is currently examining environmental impacts and adopting measures that mitigate the impacts making the process less destructive to the environment.

With the current commitment to sustainability, opal companies are investing funds for Green Mining as a positive way to impact the environment before and after mining.


7 0
4 years ago
Read 2 more answers
According to universal gravitation, both mass and air resistance affect the gravitational attraction between objects
Ksju [112]
Air resistance doesn't appear in the formula for gravitational force, because it doesn't affect it. Mass does because it does.
7 0
3 years ago
Read 2 more answers
If it takes Ashley 10 seconds to run from the batter's box to first base at an average speed
abruzzese [7]

Answer:

65 meters

Explanation:

Time = 10 seconds

Speed = 6.5 m/s

Distance = ?

Distance = speed\times time\\\\d = 6.5 \times 10 \\d = 65 m

7 0
3 years ago
A jogger travels a route that has two parts. The first is a displacement of 3 km due south, and the second involves a displaceme
spayn [35]

Answer:

a) 2.41 km

b) 38.8°

Questions c and d are illegible.

Explanation:

We can express the displacements as vectors with origin on the point he started (0, 0).

When he traveled south he moved to (-3, 0).

When he moved east he moved to (-3, x)

The magnitude of the total displacement is found with Pythagoras theorem:

d^2 = dx^2 + dy^2

Rearranging:

dy^2 = d^2 - dx^2

dy = \sqrt{d^2 - dx^2}

dy = \sqrt{3.85^2 - 3^2}  = 2.41 km

The angle of the displacement vector is:

cos(a) = dx/d

a = arccos(dx/d)

a = arccos(3/3.85) = 38.8°

7 0
3 years ago
A ball is projected with an initial velocity of 40 meter per second and reached maximum height of 160 meters calculate tge angle
Andru [333]

There's a problem with the question as given. Even with a maximum projection angle of <em>θ</em> = 90°, the initial velocity is not large enough to get the ball up in the air 160 m. With angle 90°, the ball's height <em>y</em> at time <em>t</em> would be

<em>y</em> = (40 m/s) <em>t</em> - 1/2 <em>g t</em> ²

Set <em>y</em> = 160 m, and you'll find that there is no (real) solution for<em> t</em>, so the ball never attains the given maximum height.

From another perspective: recall that

<em>v </em>² - <em>v</em>₀² = 2<em>a </em>∆<em>y</em>

where

• <em>v</em>₀ = initial velocity

• <em>v</em> = final velocity

• <em>a</em> = acceleration

• ∆<em>y</em> = displacement

At its maximum height, the ball has zero vertical velocity, and ∆<em>y</em> = maximum height = 160 m. The ball is in free fall once it's launched, so <em>a</em> = -<em>g</em>.

So we have

0² - (40 m/s)² = -2<em>g </em>(160 m)

but this reduces to

(40 m/s)² = 2 (9.8 m/s²) (160 m)

1600 m²/s² ≠ 3136 m²/s²

7 0
3 years ago
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