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sdas [7]
3 years ago
6

Dolphins communicate using compression waves (longitudinal waves). Some of the sounds dolphins make are outside the range of hum

an hearing. Upon spotting a hungry shark, dolphin A sends a message to dolphin B. To send this message as effectively as possible, dolphin A would Question 1 options: send the message above the water when it jumps because air is more dense than water, and the sound would be louder above the water. make a higher pitched signal because a higher pitch is a louder sound. send the message underwater because a more dense medium would make the sound travel faster. make a softer sound because the shark will not hear the softer signal.
Physics
1 answer:
lana66690 [7]3 years ago
3 0

Answer: send the message underwater because a more dense medium would make the sound travel faster.

Explanation:

Dolphins communicate using compression waves - longitudinal waves. Longitudinal waves requires a medium to travel. A longitudinal wave transfers energy by the vibration of medium particles in the direction of the wave motion. Compression are the regions where density of the medium is higher and rarefaction is a low density region.

A longitudinal wave travels faster in a denser medium. It has maximum speed in solid and minimum in gas. Thus, to transfer message quickly to dolphin B., dolphin A should send the message underwater and not in air. This is because water has higher density than air. Molecules collide more quickly in water than in air and it takes less time for signal to travel.

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Infer​ the effect of deforestation on the carrying capacity of the Amazon rainforest.
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Answer:

More than 20% of the amazon has been destroyed, affects biodiversity.

Explanation:

  • The biggest issues that the amazon is facing is the deforestation and it involves the clearing of land areas for logging, farming, and other land-use changes.
  • Effects that are seen are droughts, floods, destruction of habitats of flora and fauna along with the valuable services of the ecosystem.
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4 0
3 years ago
The equation for the speed of an electromagnetic wave is
igomit [66]

Answer:

c = 1 / √(ε₀*μ₀)

Explanation:

The speed of the electromagnetic wave in free space is given in terms of the permeability and the permittivity of free space by

c = 1 / √(ε₀*μ₀)

where the permeability of free space (μ₀) is a physical constant used often in electromagnetism and ε₀ is the permittivity of free space (a physical constant).

4 0
3 years ago
A charge of 31.0 μC is to be split into two parts that are then separated by 24.0 mm, what is the maximum possible magnitude of
miskamm [114]

Answer:

1.72 x 10³ N.

Explanation:

When a charge is split equally and placed at a certain distance , maximum electrostatic force is possible.

So the charges will be each equal to

31/2 = 15.5 x 10⁻⁶ C

F = K Q q / r²

= \frac{9\times10^9\times(10.5)^2\times10^{-12}}{(24\times10^{-3})^2}

= 1.72 x 10³ N.

8 0
3 years ago
Linear expansivity?<br>​
shusha [124]

Linear expansivity, area expansivity and volume or cubic expansivity are

7 0
2 years ago
Example 3 :
kherson [118]

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

μ = viscosity = 0. 06 Pas

d =  diameter = 120mm = 0. 12m

V =  velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × \frac{0. 06}{0. 12* 1* 0. 9}

friction factor = 0. 52 × \frac{0. 06}{0. 108}

friction factor = 0. 52 × 0. 55

friction factor = 0. 289

b. When V = 3mls

Friction factor = 0. 52 × \frac{0. 06}{0. 12 * 3* 0. 9}

Friction factor = 0. 52 × \frac{0. 06}{0. 324}

Friction factor = 0. 52 × 0. 185

Friction factor = 0.096

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × \frac{1}{2*6. 6743 * 10^-11} × \frac{1^2}{0. 120} × \frac{1}{100}

Head loss =  1. 80 × 10^8

Head loss When V = 3m/s

Head loss = 0. 096 × \frac{1}{1. 334 *10^-10} × \frac{3^2}{0. 120} × \frac{1}{100}

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss  when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

Learn more about friction here:

brainly.com/question/24338873

#SPJ1

4 0
1 year ago
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