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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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The answer is A. Sees more intense wavelengths reaching the surface
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2 years ago
A battery has an emf of 15.0 V. The terminal voltage of the battery is 12.2 V when it is delivering 14.0 W of power to an extern
solong [7]

Answer:

The value of R and the internal resistance of the battery are 10.6 ohm and 2.45 ohm

Explanation:

Given that,

Emf of battery = 15.0 V

Voltage = 12.2 V

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(a). We need to calculate the value of R

Using formula of power

P=IV

P=\dfrac{V^2}{R}

R=\dfrac{V^2}{P}

Where, R = resistance

P = power

V = voltage

Put the value into the formula

R =\dfrac{(12.2)^2}{14.0}

R=10.6\ \Omega

(b). We need to calculate the internal resistance of the battery

Firstly we calculate the current

Using formula of current

I=\dfrac{V}{R}=\dfrac{P}{V}

Put the value of P and V into the formula

I =\dfrac{14}{12.2}

I=1.14\ A

We calculate the internal resistance

Using formula of emf

e-V=Ir

r =\dfrac{e-V}{I}

Put the value into the formula

r=\dfrac{15.0-12.2}{1.14}

r = 2.45\ \Omega

Hence, The value of R and the internal resistance of the battery are 10.6 ohm and 2.45 ohm

3 0
3 years ago
The wavelengths of visible light vary from about 300 nm to 700 nm. What is the range of frequencies of visible light in a vacuum
exis [7]

The range of frequencies  of visible light in a vacuum is mathematically given as

Fmin=4.19*10^14Hz  to Fmax=1*10^15Hz

<h3>What is the range of frequencies of visible light in a vacuum?</h3>

Question Parameters:

The wavelengths of visible light vary from about 300 nm to 700 nm.

Generally, the equation for the frequency   is mathematically given as

F=C/\lambda

Therefore

For Fmax

Fmax=\frac{300*10^8}{3*10^9}

Fmax=1*10^15Hz

Where

Fmin=\frac{3*10^8}{700*10^9}

Fmin=4.19*10^14Hz

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