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olchik [2.2K]
3 years ago
9

A sound wave enters a new medium where sound travels faster. How does this affect the frequency and wavelength of the sound?

Physics
2 answers:
mamaluj [8]3 years ago
8 0
<h3><u>Answer;</u></h3>

The frequency stays the same and the wavelength increases.

<h3><u>Explanation;</u></h3>
  • <em><u>Sound waves are a type of mechanical wave which requires a material medium for transmission. The transmission of such waves occurs as a result of vibration of particles in the material medium. The faster the vibration of the particles the higher the speed of the sound waves.</u></em>
  • <em><u>When a sound enters a new medium where the sound travels faster, it means the vibration of particles is high, and the wavelength of the waves increases which means the speed of the wave has increased. The frequency on the other hand will remain constant.</u></em>
iren [92.7K]3 years ago
6 0
The frequency doesn't change. If the wavespeed increases, then the wavelength must also increase ... It's just the distance the wave travels during each complete wiggle.
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100%
xxMikexx [17]

Answer:

1. The elastic potential energy is 0.0176 Joules

2. The kinetic energy of the pinball the instant it leaves the spring is 0.0176 Joules

3. The speed of the pinball the instant it leaves the spring is approximately 2.42212 m/s

4. The height of the part where the pinball is located on the machine above the ground is approximately 0.213 meters

Explanation:

The spring constant of the pinball machine's plunger, k = 22 N/m

The amount by which the pinball machine's plunger is compressed, x = 0.04 m

The mass of the pinball ball, m = 0.006 kg

1. The elastic potential energy, P.E. = 1/2·k·x²

By substitution, we get;

P.E. = 1/2 × 22 N/m × (0.04 m)² = 0.0176 J

The elastic potential energy, P.E. = 0.0176 J

2. At the instant the pinball leaves the spring, the plunger and therefore the force of the plunger no longer acts on the pinball

Since there are no external forces acting on the pinball to increase the speed of the pinball after it leaves the spring, the velocity reached is its maximum velocity, and therefore, the kinetic energy, K.E. is the maximum kinetic energy which by the conservation of energy, is equal to the initial potential energy

Therefore;

K.E. = P.E. = 0.0176 J

The kinetic energy of the pinball the instant it leaves the spring, K.E.= 0.0176 J

3. The kinetic energy, K.E., is given by the following formula;

K.E. = 1/2·m·v²

Where;

v = The speed or velocity of the object having kinetic energy K.E.

Therefore, from K.E. = 0.0176 J, and by plugging in the values of the variables, we have;

K.E. = 0.0176 J = 1/2 × 0.006 kg × v²

v² = 0.0176 J/(1/2 × 0.006 kg) = 88/15 m²/s²

v = √(88/15 m²/s²) ≈ (2·√330)/15 m/s ≈ 2.42212 m/s

The speed of the pinball the instant it leaves the spring, v ≈ 2.42212 m/s

4. The height of the pinball is given by the following kinematic equation of motion;

v_h² = u² - 2·g·h

Where;

v_h = The velocity of the pinball at the given height = 1.3 m/s

u = v ≈ 2.42212 m/s (The initial velocity of the pinball as it the spring)

g = The acceleration due to gravity ≈ 9.8 m/s²

h = The height of the pinball above the ground

We get;

v_h² = 1.3² = 2.42212² - 2 × 9.8 × h

∴  h = (2.42212² - 1.3²)/(2 × 9.8) ≈ 0.213

The height of the part where the pinball is located on the machine above the ground, h ≈ 0.213 m

5 0
3 years ago
A child in a boat throws a 5.80-kg package out horizontally with a speed of 10.0 m/s. The mass of the child is 24.6kg and the ma
Sergio [31]

Answer:

-0.912 m/s

Explanation:

When the package is thrown out, momentum is conserved. The total momentum after is the same as the total momentum before, which is 0, since the boat was initially at rest.

(m_c + m_b)v_b + m_pv_p = 0

where m_c = 24.6 kg, m_b = 39 kg, m_p = 5.8 kg are the mass of the child, the boat and the package, respectively. , v_p = 10m/s, v_b are the velocity of the package and the boat after throwing.

(24.6 + 39)v_b + 5.8*10 = 0

63.6v_b + 58 = 0

v_b = -58/63.6 = -0.912 m/s

3 0
4 years ago
Helppppppp mee........​
Ket [755]

Answer:

7) 0.0025510294

8) 100

5 0
3 years ago
A subway car moves at a constant speed of 10 m/s over a period of 10 s. What is the instantaneous speed halfway through this mot
andre [41]

Answer:  10 m/s

We're told the speed is constant, so it's not changing throughout the time period given to us. So throughout the entire interval, the speed is 10 m/s.

5 0
3 years ago
What is the frequency of oscillation for a mass on the end of spring with a period of motion of 2.6 seconds? Calculate answer to
Tcecarenko [31]

Answer:

Frequency, f = 0.38 Hz

Explanation:

Time period of the spring, T = 2.6 seconds

We need to find the frequency of oscillation for a mass on the end of spring. The relation between the time period and the frequency is given by :

Let f is the frequency of oscillation. So,

f=\dfrac{1}{T}

f=\dfrac{1}{2.6\ s}

f = 0.38 Hz

or

f = 0.4 Hz

So, the frequency of oscillation for a mass on the end of a spring is 0.38 hertz. Hence, this is the required solution.

4 0
4 years ago
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