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3 years ago

8

Asexual reproduction provides an advantage to organisms under which of the following conditions?

1 answer:

krek1111 [17]3 years ago

4 0

Answer:

The organisms are widely dispersed in the habitat and rarely encounter each other...

Explanation:

As asexual reproduction needs only 1 parent.. they can do it on their own.. without encountering their mates

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In terms of physics why should a golfer follow his or her swing

Nikitich [7]

He or she should follow her swing due to the momentum that was built

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4 years ago

State two other ways in which evaporation is different from boiling

Nookie1986 [14]

Answer:

To summarize, evaporation is slower, occurs only from the surface of the liquid, does not produce bubbles, and leads to cooling. Boiling is faster, can occur throughout the liquid, produces lots of bubbles, and does not result in cooling.

Explanation:

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3 years ago

Read 2 more answers

How much heat does it take to bring 0.12 kg of liquid water at 100c to steam at 140c

Vadim26 [7]

Answer:

162

Explanation:

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3 years ago

Two guitarists attempt to play the same note of wavelength 6.50 cm at the same time, but one of the instruments is slightly out

PolarNik [594]

Answer:

The two value of the wavelength for the out of tune guitar is

$\lambda _2 = (6.48,6.52) \ cm$

Explanation:

From the question we are told that

The wavelength of the note is $\lambda = 6.50 \ cm = 0.065 \ m$

The difference in beat frequency is $\Delta f = 17.0 \ Hz$

Generally the frequency of the note played by the guitar that is in tune is

$f_1 = \frac{v_s}{\lambda}$

Where $v_s$ is the speed of sound with a constant value $v_s = 343 \ m/s$

$f_1 = \frac{343}{0.0065}$

$f_1 = 5276.9 \ Hz$

The difference in beat is mathematically represented as

$\Delta f = |f_1 - f_2|$

Where $f_2$ is the frequency of the sound from the out of tune guitar

$f_2 =f_1 \pm \Delta f$

substituting values

$f_2 =f_1 + \Delta f$

$f_2 = 5276.9 + 17.0$

$f_2 = 5293.9 \ Hz$

The wavelength for this frequency is

$\lambda_2 = \frac{343 }{5293.9}$

$\lambda_2 = 0.0648 \ m$

$\lambda_2 = 6.48 \ cm$

For the second value of the second frequency

$f_2 = f_1 - \Delta f$

$f_2 = 5276.9 -17$

$f_2 = 5259.9 Hz$

The wavelength for this frequency is

$\lambda _2 = \frac{343}{5259.9}$

$\lambda _2 = 0.0652 \ m$

$\lambda _2 = 6.52 \ cm$

8 0

3 years ago

If you swim with the current in a river, your speed is increased by the speed of the water; if you swim against the current, you

Blababa [14]

Answer:

11.23%

Explanation:

Lets take

Speed of man in still water =u= 1.73 m/s

Speed of flow of water = v=0.52 m/s

When swims in downward direction then speed of man = u + v

When swims in upward direction then speed of man = u - v

Lets time taken by man when he swims in downward direction is $t_1$ and when he swims in downward direction is $t_2$

Lets distance is d and it will be remain constant in both the case

$d=(u+v)t_1$

$d=(u-v)t_2$

$(1.73+0.52)t_1=(1.73-0.52)t_2$

$t_2=1.85t_1$

Time taken in still water

2 d= t x 1.73

t=1.15 x d sec

$t_1=0.44d\ sec$

$t_2=0.82d\ sec$

total time in current = 0.82 +0.44 d=1.26 d sec

So the percentage time

$percentage\ time =\dfrac{1.28-1.15}{1.15}$

Percentage time =11.32%

So it will take 11.32% more time as compare to still current.

5 0

3 years ago