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3 years ago

Can swimmers hold their breath longer that people who don't swim. what is the hypothesis?

Physics

2 answers:

NeTakaya3 years ago

8 0

A hypothesis is a statement, not a question.

<em>Can swimmers hold their breath longer that people who don't swim ?</em>

I have no idea. What do you think ? <em>The hypothesis is what you think.</em>

I see three possibilities here:

1). Swimmers can hold their breath longer that non-swimmers.

2). People who don't swim can hold their breath longer than swimmers.

3). Average swimmers and average non-swimmers can generally hold their breath for the same length of time.

These are three possible hypotheses. There may be others that I didn't think of.

What do you think ? Once you decide what YOU think, you have a hypothesis. Your next job is to set up some kind of test or experiment to find out whether your hypothesis is true or not.

stealth61 [152]3 years ago

5 0

The hypothesis would be can professional swimmer hold their breath longer than people who aren't

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yuradex [85]

Answer:

5.71428571422 m/s

Explanation:

u = Initial velocity = 20 m/s

v = Final velocity

s = Displacement

a = Acceleration

Time taken = 15-1 = 14 s

Distance traveled in 1 second = $20\times 1=20\ m$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 200-20=20\times 14+\frac{1}{2}\times a\times 14^2\\\Rightarrow a=\frac{2(180-20\times14)}{14^2}\\\Rightarrow a=-1.02040816327\ m/s^2$

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3 years ago

A motorcycle traveling at 25 m/s accelerates at a rate of 6 m/s2 for 4 seconds. What is the final speed of the motorcycle in m/s

givi [52]

Initial velocity = Vo= 25 m/s

Final velocity = V = x

Acceleration= a = 6 m/s^2

time= t = 4 seconds

Appy the equation:

V = Vo + at

Replacing:

V = 25 + 6(4) = 25 + 24 = 49 m/s

8 0

1 year ago

Light with an intensity of 1 kW/m2 falls normally on a surface and is completely absorbed. The radiation pressure is

kobusy [5.1K]

Answer:

The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

Explanation:

Given;

intensity of light, I = 1 kW/m²

The radiation pressure of light is given as;

$Radiation \ Pressure = \frac{Flux \ density}{Speed \ of \ light}$

I kW = 1000 J/s

The energy flux density = 1000 J/m².s

The speed of light = 3 x 10⁸ m/s

Thus, the radiation pressure of the light is calculated as;

$Radiation \ pressure = \frac{1000}{3*10^{8}} \\\\Radiation \ pressure =3.33*10^{-6} \ Pa$

Therefore, the radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

6 0

3 years ago

A person sitting on a pier observes incoming waves that have a sinusoidal form with a distance of 2.5 m between the crests. Of a

Doss [256]

Answer:

Part(a): The frequency is $\bf{0.2~Hz}$ .

Part(b): The speed of the wave is $\bf{0.5~m/s}$ .

Explanation:

Given:

The distance between the crests of the wave, $d = 2.5~m$ .

The time required for the wave to laps against the pier, $t = 5.0~s$

The distance between any two crests of a wave is known as the wavelength of the wave. So the wavelength of the wave is $\lambda = 2.5~m$ .

Also, the time required for the wave for each laps is the time period of oscillation and it is given by $T = 5.0~s$ .

Part(a):

The relation between the frequency and time period is given by

$\nu = \dfrac{1}{T}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Substituting the value of $T$ in equation (1), we have

$\nu &=& \dfrac{1}{5.0~s}\\~~~&=& 0.2~Hz$

Part(b):

The relation between the velocity of a wave to its frequency is given by

$v = \nu \lambda~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Substituting the value of $\nu$ and $\lambda$ in equation (2), we have

$v &=& (0.2~Hz)(2.5~m)\\~~~&=& 0.5~m/s$

5 0

3 years ago

run at the same speed and in the same direction, and they both run for the same amount of time, what can you say about the dista

LenaWriter [7]

Answer:

the are equivalent

Explanation:

i just learned about that

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4 years ago