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nataly862011 [7]
3 years ago
13

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:

make a softer sound because the shark will not hear the softer signal.


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.


send the message underwater because a more dense medium produces a louder sound.


make a higher pitched signal because a higher pitch is a louder sound.
Physics
1 answer:
Sphinxa [80]3 years ago
8 0
Water is a really good conductor of sound so I would have to say that it would be to send the message underwater because a more dense medium produces a louder sound
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Two train 75 km apart approach each other on parallel tracks, each moving at
creativ13 [48]

Answer:

The correct solution is "37.5 km".

Explanation:

Given:

Distance between the trains,  

d = 75 km

Speed of each train,

= 15 km/h

The relative speed will be:

= 15 + (-15)

= 30 \ km/h

The speed of the bird,

V = 15 km/h

Now,

The time taken to meet will be:

t=\frac{Distance}{Relative \ speed}

  =\frac{75}{30}

  =2.5 \ h

hence,

The distance travelled by the bird in 2.5 h will be:

⇒ D = V t

        =15\times 2.5

        =37.5 \ km

 

6 0
3 years ago
What statement is NOT true about standard electrode potentials?
creativ13 [48]

Answer:

(a) E^{\circ}_{cell}  is the difference in voltage between the anode and the cathode .

Explanation:

If E^{\circ}_{cell} will be the difference anode and the cathode means E^{\circ}_{cell}=E^{\circ}_{anode}-E^{\circ}_{cathode}n as we know that cathode potential is greater than the anode potential so E^{\circ}_{cell} will become negative and if E^{\circ}_{cell}  is negative then reaction will not be spontaneous so the correct relation of  E^{\circ}_{cell} is E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}

4 0
4 years ago
Suppose you have a polarized light source in your lab that has an intensity of 1288 W/m2, where the electric field oscillates in
IRINA_888 [86]

Answer:

644 W/m²

0 W/m²

Explanation:

I_0 = Initial intensity of light = 1288 W/m²

\theta = Angle of filter

Intensity of light is given by

I=I_0cos^2\theta\\\Rightarrow I=1288\times cos^2(45)\\\Rightarrow I=644\ W/m^2

The intensity of the transmitted light is 644 W/m²

I=I_0cos^2\theta\\\Rightarrow I=1288\times cos^2(90)\\\Rightarrow I=0\ W/m^2

The intensity of the transmitted light is 0 W/m²

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3 years ago
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Explanation:

The answer is mentioned above.

Hope it helps.

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3 years ago
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