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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.

Can swimmers hold their breath longer that people who don't swim ?

I have no idea. What do you think ? The hypothesis is what you think.

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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Find the useful power output (in W) of an elevator motor that lifts a 2600 kg load a height of 30.0 m in 12.0 s, if it also incr

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Answer:

P = 251,916.667 W

Cost = 2,267.25 cents

Explanation:

To solve this question we will use the Work Energy Theorem, which is

$W = dP + dK\\$

Where

$dP$ = Change in Potential Energy

$dK$ = Change in Kinetic Energy

Change in Potential Energy

$P_{i} = mgh_{i}\\ P_{f} = mgh_{f}$

Where

$P_{i}$ = Initial Potential Energy

$P_{f}$ = Final Potential Energy

$m$ = Mass of System = 10,000 kg

$g$ = Acceleration due to gravity = 9.81 m/s

$h_{i}$ = Initial Height = 0

$h_{f}$ = Final Height = 30 m

Inputting the values we get the answer for $dP$

$dP = P_{f} - P_{i}\\dP= mgh_{f} - mgh_{i}\\ dP= 10000(9.81)(30) - 0\\ dP= 2943000$

Change in Kinetic Energy

$K_{i} = \frac{1}{2} mv_{i} ^2\\ K_{f} = \frac{1}{2} mv_{f} ^2$

Where

$K_{i}$ = Initial Kinetic Energy

$K_{f}$ = Final Kinetic Energy

$m$ = Mass of System = 10,000 kg

$g$ = Acceleration due to gravity = 9.81 m/s

$v_{i}$ = Initial Velocity = 0 m/2

$v_{f}$ = Final Velocity = 4 m/s

Inputting the values we get the answer for $dK$

$dK = K_{f} - K_{i}\\ dK = \frac{1}{2} mv_{f} ^2 - \frac{1}{2} mv_{i} ^2\\ dK = \frac{1}{2} (10000)(4)^2 - 0 \\ dK = 80000$

Total Work

$W = dP + dK\\$

Inputting the values

$W = 2943000 + 80000$

$W = 3,023,000$

a) Finding the useful Power Output

$P = \frac{W}{t}$

Where

$P$ = Power Output

$W$ = Work Done = 3,023,000J

$t$ = Time = 12s

Inputting the values

$P = \frac{3,023,000}{12}\\ P = 251,916.667$

P = 251,916.667 W

b) Finding the Total Cost

Cost = $0.0900 x P/1000

Cost = $0.0900 x (251,916.667/1000)

Cost = $22.67 or 2,267.25 cents

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