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Dennis_Churaev [7]
3 years ago
15

Humans can typically perceive frequencies that range from 20 Hz–20 kHz. Dogs can hear sounds up to 45 kHz. Dog whistles produce

sound waves that range from 23–54 kHz; therefore, they cannot be heard by the human ear.
Which statement accurately describes the pitch and wavelength of the sound waves produced by a dog whistle?

high pitch with a long wavelength
high pitch with a short wavelength
low pitch with a short wavelength
low pitch with a long wavelength
Physics
2 answers:
Aleksandr [31]3 years ago
6 0
Pitch is directly related to the frequency of the sound. In this item, we are given that the frequency of the sound is higher compared to those which are audible to the human being's ears. The pitch therefore of the dog's whistle is high. 

On the other hand, the frequency and the wavelength of a certain wave are inversely proportional. This means that the high frequency wave will have a short wavelength. 

Hence, the answer to this item would have to be "high pitch with a short wavelength" 

The answer to this item is the second option. 
Svetach [21]3 years ago
6 0
High pitch with a short wavelength 

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Two velcro-covered pucks slide across the ice, collide and stick to one another. Their interaction with the ice is frictionless.
balu736 [363]

Answer:

<em>1. False</em>

<em>2. True</em>

<em>3. False</em>

<em>4. True</em>

Explanation:

<u>Conservation of Momentum</u>

According to the law of conservation of linear momentum, the total momentum of the system formed by both pucks won't change regardless of their interaction if no external forces are acting on the system.

The momentum of an object of mass ma moving at speed va is

p_a=m_a.v_a

The total momentum of both pucks at the initial condition is

p_1=m_a.v_a+m_b.v_b

Both pucks are moving to the right and puck B has twice the mass of puck A (let's call it m), thus

m_a=m

m_b=2m

We are given

v_a=6\ m/s\\v_b=2\ m/s

The total initial momentum is

p_1=6m+2(2m)=10m

At the final condition, both pucks stick together, thus the total mass is 3m and the final speed is common, thus

p_2=3m.v'

Equating the initial and final momentum

10m=3m.v'

Solving for v'

v'=10/3\ m/s=3.33\ m/s

1. Compute the initial kinetic energy:

\displaystyle K_1=\frac{1}{2}mv_a^2+\frac{1}{2}2mv_b^2

\displaystyle K_1=\frac{1}{2}m\cdot 6^2+\frac{1}{2}2m\cdot 2^2

K_1=18m+4m=22m

The final kinetic energy is

\displaystyle K_2=\frac{1}{2}mv'^2+\frac{1}{2}2mv'^2

\displaystyle K_2=\frac{1}{2}m\cdot 3.33^2+\frac{1}{2}2m\cdot 3.33^2

K_2=16.63m

As seen, part of the kinetic energy is lost in the collision, thus the statement is False

2. The initial speed of puck B was 2 m/s and the final speed was 3.33 m/s, thus it increased the speed: True

3. The initial speed of puck A was 6 m/s and the final speed was 3.33 m/s, thus it decreased the speed: False

4. The momentum is conserved since that was the initial assumption to make all the calculations. True

p_1=10m

p_2=3m.v'=3m(10/3)=10m

Proven

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3 years ago
What is the Kinetic Energy of a 60 kg person on skis traveling 20 m/s?
sattari [20]

Explanation:

equating the parameters into the formula, it's gonna be

= ½ × 60 × 20²

= ½ × 60 × 400

= ½ × 24000

K.E = 12000J

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3 years ago
Explain Archimedes Principle
rusak2 [61]

Answer:

Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially, is equal to the weight of the fluid that the body displaces. Archimedes' principle is a law of physics fundamental to fluid mechanics. It was formulated by Archimedes of Syracuse.

Explanation:

6 0
2 years ago
12.51 A parallel RLC circuit, which is driven by a variable frequency 2-A current source, has the following values: R = 1 kΩ, L
Anastaziya [24]

Answer:

BW = 100 rad/s

wlow = 452.49 rad/s

whigh = 552.49 rad/s

V(jwlow) =1414.21 < 45°V

V(jwhigh) =1414.21 <-45°V

Explanation:

To calculate bandwidth we have formula

BW = 1/RC

BW = 1/ 1000x10x10^¯6

BW = 100 rad/s

We will first calculate resonant frequency and quality factor for half power frequencies.

For resonant frequency

wo = 1/(SQRT LC)

wo = 1/SQRT 400×10¯³ × 10×10^¯6

wo = 500 rad/s

For Quality

Q = wo / BW

Q = 500/100

Q = 5

wlow = wo [-1/2Q+ SQRT (1/2Q)² + 1]

wlow = 500 [-1/2×5 + SQRT (1/2×5)² + 1]

wlow = 452.49 rad/s

whigh = wo [1/2Q+ SQRT (1/2Q)² + 1]

whigh = 500 [1/2×5 + SQRT (1/2×5)² + 1]

whigh = 552.49 rad/s

We will start with admittance at lower half power frequency

Y(jwlow) = (1/R) + (1/jwlow L) + (jwlow C)

Y(jwlow) = (1/1000) + (1/j×452.49×400×10¯³) + (j×452.49×10×10^¯6)

Y(jwlow) = 0.001 - j5.525×10¯³ + j4.525×10¯³

Y(jwlow) = (1-j).10¯³ S

Voltage across the network is calculated by ohm's law

V(jwlow) = I/Y(jwlow)

V(jwlow) = 2/(1-j).10¯³

V(jwlow) = 1414.2 < 45°V

Now we will calculate the admittance at higher half power frequency

Y(jwhigh) = (1/R) + (1/jwhigh L) + (jwhigh C)

Y(jwhigh) = (1/1000) + (1/j×552.49×400×10¯³) + (j×552.49×10×10^¯6)

Y(jwhigh) = 0.001 - j4.525×10¯³ + j5.525×10¯³

Y(jwhigh) = (1+j).10¯³ S

Voltage across network will be calculated by ohm's law

V(jwhigh) = I/Y(jwhigh)

V(jwhigh) = 2/(1+j).10¯³

V(jwhigh) = 1414.2 < - 45°V

6 0
3 years ago
A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave persp
My name is Ann [436]

Answer:

915m

Hope this helps.

5 0
3 years ago
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