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motikmotik
2 years ago
10

Which statement best describes the motion of air particles in a sound wave?

Physics
2 answers:
Elodia [21]2 years ago
8 0

Answer:

The answer is D.

Explanation:

They vibrate parallel to the wave.

During the propagation of a sound wave in air, the vibrations of the particles are most accurately represented as longitudinal. Longitudinal waves are waves in which the motion of the individual particles of the medium occurs in a direction that is parallel to the direction of energy transmission.

Zolol [24]2 years ago
7 0
Answer
D
Mark me brilliant
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The following data were collected during a short race between two friends. Velocity (m/s) 0 0.5 1 1.5 2 2 4 6 2 0 Time (s) 0 2 4
scoundrel [369]

The characteristics of the kinematics allow to find the results for the questions about the movement of the body are:

a)  we have four sections;

  • 0 to 8 s The body is accelerating.
  • 8 to 10 s The body goes at a constant speed, the acceleration is zero.
  • 10 to 14 Body accelerating.
  • 14 to 18 Body slowing down.

b)  The acceleration is the first 8 s is:  a = 0.25 m / s²

c) The maximum acceleration is:    a = 1 m / s²

d) The displacement   is:  i) d₁ =  8m,     ii)  d_{total}= 16 m

e) maximum speed  is:      v = 6 m / s

Kinematics studies the movement of bodies by finding relationships between the position, speed and acceleration of bodies.

        v = v₀ + a t

        y = v₀ t + ½ a t²

where v and v₀ is the current and initial velocity, respectively, a is the acceleration and t is time.

In many circumstances graphs are made for their analysis, in a graph of speed versus time when we have a horizontal line the speed is constant, the acceleration is zero and in the case of a slope there is an acceleration, we have two cases:

  • Positive slope the body is accelerating and the speed is increasing.
  • Negative slope the body is stopping, the speed decreases.

Let's answer the different questions about the system.

a) in the attached we have a graph of the velocity versus time, each section corresponds to a change in the slope of the graph, we have four sections;

  • 0 to 8 s The body is accelerating.
  • 8 to 10 s The body goes at a constant speed, the acceleration is zero.
  • 10 to 14 Body accelerating.
  • 14 to 18 Body slowing down.

b) The acceleration is the first 8 s

          v = v₀ + a t

          a = \frac{v-v_o}{\Delta t}  

          a = \frac{2-0}{8-0}  

          a = 0.25 m / s²

c) The maximum acceleration is when the slope is maximum.

          a = \frac{6-2}{ 14-10}  

          a = 1 m / s²

Therefore the acceleration is maximum in the section between 10 and 14 s

d) The total displacement is the sum of the displacements of each section.

         d_{total } = d_1 +d_2 + d_3 +d_4  

We look for every displacement.

       d₁ = v₀ + ½ a₁ Δt²

       d₁ = 0 + ½ 0.25 8²

       d₁ = 8 m

In the second section the velocity is constant

         d₂ = v₂ Δt₂

         d₂ = 2 (10-8)

         d₂ = 4 m

The third section.

    d₃ = v₀ + ½ a t²

    d₃ = 2 + ½ 1 (14-10) ²

    d₃ = 10 m

The distance of the fourth section.

       

we look for acceleration

          a₄ = \frac{v-v_o}{\Delta t}  

          a₄ = \frac{0-6}{18-14}  

          a₄ = -1.5 m / s²

     

          d₄ = 6 + ½ (-1.5) (1814) ²

          d₄ = -6 m

The total displacement is;

          d_{total} = 8 + 4 + 10 -6

          d_{total} = 16 m

e) The maximum speed is the highest point in the graph of speed versus time that in the attachment we can see corresponds to

          v = 6 m / s

In conclusion using the characteristics of kinematics we can find the results for the questions about the motion of bodies are:

  a)  we have four sections;

  • 0 to 8 s The body is accelerating.
  • 8 to 10 s The body goes at a constant speed, the acceleration is zero.
  • 10 to 14 Body accelerating.
  • 14 to 18 Body slowing down.

b)  The acceleration is the first 8 s is:  a = 0.25 m / s²

c) The maximum acceleration is:    a = 1 m / s²

d) The displacement   is:  i) d₁ =  8m,     ii)  d_{total}= 16 m

e) maximum speed  is:      v = 6 m / s

Learn more about kinematics here: brainly.com/question/24783036

3 0
2 years ago
What is the velocity in meters per second of a runner who runs exactly 110 m toward the beah in 72 seconds?
rjkz [21]
B) 1.53 m/s towards the beach
8 0
3 years ago
Read 2 more answers
Marx argued that what happens when a worker is separated
kicyunya [14]

Answer:

well its simple the worker wouldn't actually be working

3 0
3 years ago
(15pts) A hungry 12.0 kg fish is coasting from west to east at 75 cm/s when it suddenly swallows a 1 kg fish swimming towards it
faust18 [17]

Answer:

The speed of the big fish after swallowing the small fish is 0.38 m/s.

Explanation:

Consider west to east direction as positive and the opposite direction as negative.

Given:

Mass of big fish (m₁) = 12.0 kg

Initial velocity of big fish (u₁) = 75 cm/s = 0.75 m/s

Mass of small fish (m₂) = 1 kg

Initial velocity of small fish (u₂) = -4 m/s (Direction is opposite to u₁)

After swallowing the small fish, both the fishes move together with same velocity. Let the velocity be 'v'.

So, as there are no effects of drag or any other forces, the given scenario can be considered as a case of inelastic collision where the objects move together with same velocity after collision.

The momentum is conserved in inelastic collision. Therefore,

Initial momentum of the fishes = Final momentum of the fishes

m_1u_1+m_2u_2=(m_1+m_2)v\\\\v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}

Now, plug in the given values and solve for 'v'. This gives,

v=\frac{12.0\times 0.75+1\times (-4)}{12.0+1}\\\\v=\frac{9-4}{13}\\\\v=\frac{5}{13}=0.38\ m/s

Therefore, the speed of the big fish after swallowing the small fish is 0.38 m/s

3 0
3 years ago
A wave with a high frequency generally has a _____.
il63 [147K]
High frequency = D, short wavelength
4 0
3 years ago
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