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Zielflug [23.3K]
2 years ago
10

Why do most amphibians return to the water to reproduce?

Physics
2 answers:
Ymorist [56]2 years ago
5 0
C. amphibian eggs do not contain a protective shell
butalik [34]2 years ago
3 0
C, amphibian eggs do not contain a protective shell
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Two different simple harmonic oscillators have the same natural frequency (f=8.80 Hz) when they are on the surface of the Earth.
bekas [8.4K]

Answer:

8.80 Hz

Explanation:

The frequency of a loaded spring is given by

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

where k and m are the spring constant and the mass of the load respectively. The values of these do not change because they are internal properties of the components of the system.

Hence, the frequency of the vertical spring mass does not change and is 8.80 Hz.

On the other hand, the frequency of the simple pendulum is affected because it is given by

\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}

where g and l are acceleration due to gravity and length of the pendulum, respectively. It is thus seen that it depends on g, which changes with location. In fact, the new frequency is given by

f_2 = 8.80\sqrt{\dfrac{1.67}{9.81}}=3.63 \text{ Hz}

3 0
3 years ago
A man jogs at a speed of 1.6 m/s. His dog waits 1.8 s and then takes off running at a speed of 3 m/s to catch the man. How far w
inessss [21]

Answer:

The dog catches up with the man 6.1714m later.

Explanation:

The first thing to take into account is the speed formula. It is v=\frac{d}{t}, where v is speed, d is distance and t is time. From this formula, we can get the distance formula by finding d, it is d=v\cdot t

Now, the distance equation for the man would be:

d_{man}=v_{man}\cdot t=1.6\cdot t

The distance equation for the dog would be obtained by the same way with just a little detail. The dog takes off running 1.8s after the man did. So, in the equation we must subtract 1.8 from t.

d_{dog}=v_{dog}\cdot (t-1.8)=3\cdot (t-1.8)

For a better understanding, at t=1.8 the dog must be in d=0. Let's verify:

d_{dog}=v_{dog}\cdot (1.8-1.8)=3\cdot (0)=0

Now, for finding how far they have each traveled when the dog catches up with the man we must match the equations of each one.

d_{man}=d_{dog}

1.6\cdot t=3\cdot (t-1.8)

1.6\cdot t=3\cdot t-5.4

1.4\cdot t=5.4

t=\frac{5.4}{1.4}

t=3.8571s

The result obtained previously means that the dog catches up with the man 3.8571s after the man started running.

That value is used in the man's distance equation.

d_{man}=1.6\cdot t=1.6\cdot (3.8571)

d_{man}=6.1714m

Finally, the dog catches up with the man 6.1714m later.

6 0
3 years ago
The sound level of one person talking at a certain distance from you is 61 dB. If she is joined by 5 more friends, and all of th
Lady_Fox [76]

Answer:

Explanation:

For sound level in decibel scale the relation is

dB = 10 log I / I₀ where I₀ = 10⁻¹² and I is intensity of sound whose decibel scale is to be calculated .

Putting the given values

61 = 10 log I / 10⁻¹²

log I / 10⁻¹² = 6.1

I = 10⁻¹² x 10⁶°¹

=10^{-5.9}

intensity of sound of 5 persons

I=5\times 10^{-5.9}

dB=10log\frac{5 X 10^{-5.9}}{10^{-12}}

= 10log 5 x 10⁶°¹

= 10( 6.1 + log 5 )

= 67.98

sound level will be 67.98 dB .

4 0
3 years ago
This important factor of survival for the coral reef is
inna [77]

Answer: biotic

Explanation:

6 0
3 years ago
What is 0.94kg divided by 2.4n
LuckyWell [14K]

Answer:

0.39166666666

Explanation:

7 0
3 years ago
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