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EleoNora [17]
2 years ago
9

Which statement is true for a sound wave entering an area of warmer air?

Physics
2 answers:
Georgia [21]2 years ago
5 0

Answer:

E. The wave travels faster and with an increased wavelength.

Explanation:

As we know speed of the sound depends on the temperature of medium, higher the speed of sound more will be the temperature.

As the sound enters an area of warmer temperature, that molecules of that area posses high kinetic energy hence they vibrate faster resulting an increase in speed of sound. Now more the speed, more will be the wavelength.

Hence when sound wave enters an area of warmer air it will travel with a faster speed  and with an increased wavelength.

kherson [118]2 years ago
4 0
The appropriate response is letter D. The wave ventures slower and with an expanded wavelength when a sound wave entering a range of hotter air. Hotter air implies less thick, so the wave ought to back off.

I hope the answer will help you, feel free to ask more in brainly. 
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How do you find the force of a ping-pong ball rolling down a track?
Kaylis [27]

Answer:

Weight

a) weight's vertical component = Normal upward force

b) weight's horizontal component = Friction force = (mass of ball)(acceleration)

These forces depend upon the track,

1) inclined or horizontal

2) steepness.

Explanation

The force of gravity points straight down, but a ball rolling down a ramp doesn't go straight down, it follows the ramp. Therefore, only the component of the weight which points along the direction of the ball's motion can accelerate the ball.

weight's horizontal component = Friction force = (mass of ball)(acceleration)

The other component pushes the ball into the ramp, and the ramp pushes back.

If the ramp is horizontal, then the ball does not accelerate, as gravity pushes the ball into the ramp and not along the surface of the ramp. Hope this helps. Can u give me brainliest

Explanation:

3 0
2 years ago
7. A car moving at 10m/s (about 22.4 mph) crashes into a barrier and stops in 0.25 m.
Galina-37 [17]

Answer:

a) 0.05s

b) 4000N

Explanation:

a)When car is stopped its final velocity become zero

U- 10 m/s

V- 0 m/s

S - 0.25 m

t -?

S = (v+u)*t/2

0.25 =(10+0)*t/2

t = 0.05s

b) If we happened to calculate the avarage force we have to consider about acceleration

V= 0

U = 10

t = 0.05 s

a =?

V = U + at

0 = 10 -a * 0.05

a = 200 m/s2

F = m *a

= 20 * 200

= 4000N

6 0
3 years ago
Figure 10.20 in your textbook shows an energy diagram for a system with total energy E1. Suppose the system's total energy is E2
wolverine [178]

The particles can undergo small oscillations around x₂.

The given parameters;

  • <em>initial energy of the particles = E₁</em>
  • <em>final energy of the particles, E₂ = 0.33E₁</em>

The movement of the particles depends on the kinetic energy of the particles.

When kinetic energy of the particles is 100%, the particles can oscillate from x₁ to x₅.

However, when the total energy of this particles is reduced to one-third (¹/₃) or 33% of the initial energy of the particle, the oscillation of the particles will be reduced.

  • The maximum position the particle can oscillate is x₅
  • The half position the particles can oscillate is x₃

Since 33% is less than the half of the energy of the particle, the particle will oscillate between x₁ and x₂.

Thus, we can conclude that the particles can undergo small oscillations around x₂.

Learn more here:brainly.com/question/23910777

3 0
2 years ago
What are the highest energy level electrons of an atom called?
Stolb23 [73]
Electrons that are the highest energy level is called Valence Electrons
6 0
3 years ago
Block A weighs 14 lb and block B weighs 5 lb If B is moving downward with a velocity (vB)1 = 3 ft/s at t = 0, determine the velo
kolezko [41]

Velocity, va2 = 10.5 ft/s

<u>Explanation:</u>

From the figure:

Length of the cable = Sa + 2Sb = l

∴ vₐ = -2vb

Applying the principle of Impulse and momentum in x-direction

mv_x_1 + \int\limits^t_t {F_x} \, dt = mv_x_2

Limit is t1 to t2

-(\frac{14}{32.2}) (2) (3) - T = \frac{14}{32.2} v_y_2                           -(1)

Applying the principle of Impulse and momentum in y-direction

mv_y_1 + \int\limits^t_t {F_y} \, dt = mv_y_2

Limit is t1 to t2

(\frac{5}{32.2}) (3) - 2T + 3 = -\frac{5}{32.2} \frac{v_y_2}{2}                            -(2)

Solving equation (1) and (2), we obtain

T = 1.6lb

va2 = 10.5 ft/s

8 0
3 years ago
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