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shepuryov [24]
2 years ago
15

What are sound detectors?

Physics
2 answers:
Olin [163]2 years ago
7 0

Answer:

A sound detector comes in the shape of a rectangular board which comprises a microphone as well as a processing circuitry.

S_A_V [24]2 years ago
5 0

Answer:

Sound detection sensor works similarly to our Ears, having diaphragm which converts vibration into signals.

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If the speed of a particle is doubled, what happens to its kinetic energy?
Norma-Jean [14]

Answer:

it increases

Explanation:

maybe for example we have 5pets then you multiple them by two u get 10

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3 years ago
This diagram shows the process that powers stars. This process is called?
viktelen [127]
The process that powers stars is C)fusion
4 0
3 years ago
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A child rides her bike at a rate of 12.0 km/hr down the street. A squirrel suddenly runs in front of her so she applies the brak
Sedbober [7]
By v = u - at 
<span>=>8 = 12 - a x 0.25 </span>
<span>=>a = 4/0.25 km/hr/sec </span>
<span>=>a = 16km/hr/sec

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3 years ago
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Use the graph below to answer the following question: What is happening to the object's velocity?
hjlf

The object's velocity is decreasing.

Explanation:

From graph is the attached image, we can clearly point that the velocity of this motion is decreasing with time.

Velocity is a vector quantity.

  • The y-axis represent displacement.
  • The x-axis depicts time
  • Using the graph, we know that the slope of the line on the graph gives us the velocity as it denotes the change of displacement with time.
  • When we find the slope, it will give us a negative value which shows that the body is slowing down and not increasing speed.

learn more:

Velocity brainly.com/question/4460262

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4 0
3 years ago
A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo
zloy xaker [14]

Solution :

Given :

Mass attached to the spring = 4 kg

Mass dropped = 6 kg

Force constant = 100 N/m

Initial amplitude = 2 m

Therefore,

a). $v_{initial} = A w$

          $= 2 \times \sqrt{\frac{100}{4}}$

          = 10 m/s

Final velocity, v at equilibrium position, v = 5 m/s

Now, $\frac{1}{2}(4+4)5^2 = \frac{1}{2} kA'$

A' = amplitude = 1.4142 m

b). $T=2 \pi \sqrt{\frac{m}{k}}$

    m' = 2m

    Hence, $T'=\sqrt2 T$

c). $\frac{\frac{1}{2}(4+4)5^2 + \frac{1}{2}\times 4 \times 10^2}{\frac{1}{2} \times 4 \times 10^2}$

  $=\frac{1}{2}$

Therefore, factor $=\frac{1}{2}$

Thus, the energy will change half times as the result of the collision.

7 0
3 years ago
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