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Leto [7]
3 years ago
5

Describe an environment in which the sound of a ringing cell phone could not be transmitted, and explain why the ringing wouldn'

t be heard. Assume that the phone is working and that the cell signal has not been interrupted
Physics
1 answer:
Sveta_85 [38]3 years ago
8 0

Answer:

In a vacuum

Explanation:

Sound is a type of mechanical waves. Mechanical waves are waves that propagate through the oscillation of the particles in a medium, which can be either gas, liquid or solid.

A sound wave in air, for instance, is simply produced by the oscillations of the air particles back and forth along the direction of motion of the wave.

Given this definition, it is clear that mechanical waves (and so, sound waves as well) cannot be transmitted if there is no medium: therefore, they cannot be transmitted in a vacuum. So, the sound of the ringing bell would not be present in a vacuum.

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Zina [86]
D is the answer
Imagine the magnetic field, it emanates from both

4 0
3 years ago
Terry can ride 30 miles in 2 hours. If his riding speed is
Anni [7]

Answer:  20.4 miles

Explanation:

Here we need to use the equation:

Velocity = Distance/Time.

Initially we have that he can travel 30 miles in 2 hours, so the velocity is:

V = 30mi/2h = 15mph

Now, we reduce the velocity by 3 mph, so the new velocity is 15mph - 3 mph = 12mph.

Now we want to know the distance traveled in 1.7 hours with this velocity, this is.

Velocity*Time = Distance

12mi/h*1.7h = 20.4 miles

7 0
2 years ago
A can of sardines is made to move along an x axis from x = 0.47 m to x = 1.20 m by a force with a magnitude given by F = exp(–8x
sattari [20]
If the force were constant or increasing, we could guess that the speed of the sardines is increasing. Since the force is decreasing but staying in contact with the can, we know that the can is slowing down, so there must be friction involved.
Work is the integral of (force x distance) over the distance, which is just the area under the distance/force graph.
The integral of exp(-8x) dx that we need is (-1/8)exp(-8x) evaluated from 0.47 to 1.20 .

I get 0.00291 of a Joule ... seems like a very suspicious solution, but for an exponential integral at a cost of 5 measly points, what can you expect. On the other hand, it's not really too unreasonable. The force is only 0.023 Newton at the beginning, and 0.000067 newton at the end, and the distance is only about 0.7 meter, so there certainly isn't a lot of work going on. The main question we're left with after all of this is: Why sardines ? ?
6 0
2 years ago
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
3 years ago
Read 2 more answers
Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 1.6×104 m/s when at a distance
xz_007 [3.2K]

Answer:

v₂ = 7.6 x 10⁴ m/s

Explanation:

given,

speed of comet(v₁) = 1.6 x 10⁴ m/s

distance (d₁)= 2.7 x 10¹¹ m

to find the speed when he is at distance of(d₂) 4.8 × 10¹⁰ m

v₂ = ?

speed of planet can be determine using conservation of energy

K.E₁ + P.E₁ = K.E₂ + P.E₂

\dfrac{1}{2}mv_1^2-\dfrac{GMm}{r_1} = \dfrac{1}{2}mv_2^2-\dfrac{GMm}{r_2}

\dfrac{1}{2}v_1^2-\dfrac{GM}{r_1} = \dfrac{1}{2}v_2^2-\dfrac{GM}{r_2}

v_2^2= v_1^2 + \dfrac{2GM}{r_2}-\dfrac{2GM}{r_1}

v_2= \sqrt{v_1^2 +2GM(\dfrac{1}{r_2}-\dfrac{1}{r_1})}

v_2= \sqrt{(1.6\times 10^4)^2 +2\times 6.67 \times 10^{-11}\times 1.99 \times 10^{30}(\dfrac{1}{4.8\times 10^{10}}-\dfrac{1}{2.7\times 10^{11}})}

v₂ = 7.6 x 10⁴ m/s

3 0
2 years ago
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