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yarga [219]
2 years ago
9

Which method of resistance training creates a consistent stress on muscles throughout the range of motion?

Biology
1 answer:
almond37 [142]2 years ago
6 0

Variable Resistance it is a Exercise that involves special weight machines that change the load throughout the range of motion so that there is a more consistent stress on the muscles.  45%b percent of body mass do muscles are made.

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2. List three sources of error that could account for the differences between your values for the enthalpy of fusion of water an
Dvinal [7]

1 trial :  nothing is given for result comparision - so we have no idea if it's a mistake.

2nd trial : The results can be compared - if varies, one may go wrong, but which one?

3rd trial : If 3rd result is different from 1st and 2nd, it is unreliable.

calculating enthalpy of fusion. M, C and m,c = mass and specific heat of calorimeter and water, n, L = mass and heat of fusion of ice; T = temperature fall.

L = (mc+MC)T/n.

c=4.18 J/gK. assuming copper calorimeter , so C=0.385 J/gK.

1. M = 409g, m = 45g. T = 22c, n = 14g

L = (45*4.18+409*0.385)*22/14 = 543.0 J/g.

2. M = 409g, m = 49g, T = 20c, n = 13g

L = (49*4.18+409*0.385)*20/13 = 557.4 J/g.

3. M = 409g, m = 54g, T = 20c, n = 14g

L = (54*4.18+409*0.385)*20/14 = 547.4 J/g.

(i) Estimate error in L from spread of 3 results.

Average L = 549.3 J/g.

squared differences average (variance) = (6.236^2+8.095^2+1.859^2)/3 = 35.96

standard deviation = 5.9964

standard error = SD/(N-1) = 5.9964/2 = 3 J/g approx.

% error = 3/547 x 100% = 0.5%.

(ii) Estimate error in L from accuracy of measurements:

error in masses = +/-0.5g

error in T = +/-0.5c

For Trial 3

M = 409g, error = 0.5g

m = 463-409, error = sqrt(0.5^2+0.5^2) = 0.5*sqrt(2)

n =(516-463)-(448-409)=14, error = 0.5*sqrt(4) = 1.0g

K = (mc+MC)=383, error = sqrt[2*(0.5*4.18)^2+(0.5*0.385)^2] = 2.962

L = K*T/n

% errors are

K: 3/383 x 100% = 0.77

T: 0.5/20 x 100% = 2.5

n: 1.0/14 x 100% = 7.14

% errors in K and T are << error in n, so ignore them.

% error in L = same as in n = 7% x 547.4 = 40

The result is (i) L= 549 +/- 3 J/g or (ii) L = 550 +/- 40 J/g.

Both are very far above  334 J/g, so there is at least one systematic error  

e.g: calorimeter may not be copper, so C is not 0.385 J/gK. (If it was polystyrene, which absorbs/ transmits little heat, the effective value of C would be very low, reducing L.)

Using +/- 40 is best.

However, the spread in the actual results is much smaller

* measurements were "fiddled" to get better results; other Trials were made but only best 3 were chosen.

<h3>Other sources of error: </h3>

L=(mc+MC)T/n is too high, so n (ice melted) may be too small, or T (temp fall) too high - why?

* we have assumed initial and final temperature of ice was 0c, it may actually have been colder, so less ice would melt -which explain small values of n

* some water might have been left in container when unmelted ice was weighed (eg clinging to ice) - again this could explain small n;

* poor insulation - heat gained from surroundings, melting more ice, increasing n - but this would reduce measured L below 334 J/g not increase it.

* calorimeter still cold from last trial when next one started, not given time to reach same temperature as water - this would reduce n.

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It is important for the cell to be able to bring necessary nutrients immedriately where it needs to and to be able to dispose of harmful byproducts of the cell process. It is equally important to be able to regulate water concentration.

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The infant begins to suck the breast. the sucking stimulates the to begin to release . in response, the releases, to initiate mi
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That can be true. Everyone (mother) is different. That usually goes away after nursing for awhile.
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Which of the following is a way that bacteria cause disease?
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Enter your immune systems it can also be in the ffood you eat

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