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3 years ago

7

Based on the solubility graph. what effect does increasing the temperature of these substances have in their solubility? does th

is suggest that these substances are solids or gases? explain
PLEASE HELP

1 answer:

Thepotemich [5.8K]3 years ago

8 0

Answer:

gases because they have a straight line and if it was solid it would have a flat line to show it melt first before the temperature increase.

Hope it helps

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B IS CORRECT

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The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have st

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Complete Question:

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm . The spring constant for the tendon is the same for both groups, $31 {\rm {N}/{mm}}$ . What is the difference in maximum stored energy between the sprinters and the nonathlethes?

Answer:

$\triangle E = 12.79 J$

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Sprinters' tendons stretch, $x_s = 43 mm = 0.043 m$

Non athletes' stretch, $x_n = 32 mm = 0.032 m$

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Maximum Energy stored in the sprinter, $E_s = 0.5kx_s^2$

Maximum energy stored in the non athletes, $E_m = 0.5kx_n^2$

Difference in maximum stored energy between the sprinters and the non-athlethes:

$\triangle E = E_s - E_n = 0.5k(x_s^2 - x_n^2)\\\triangle E = 0.5*3100* (0.043^2 - 0.032^2)\\\triangle E = 0.5*31000*0.000825\\\triangle E = 12.79 J$

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3 years ago

A scaffold of mass 57 kg and length 6.5 m is supported in a horizontal position by a vertical cable at each end. A window washer

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Explanation:

The given data is as follows.

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$l_{1}$ = 6.5 m, $l_{2}$ = (6.5 - 1.9) m = 4.6 m

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$m_{1}g \frac{l_{1}}{2} + m_{2}g \times l_{2} - T \times l_{1}$ = 0

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T = 828 N

Therefore, 828 N is the tension in the cable closer to the painter.

(b) Now, we will calculate the sum about close ends as follows.

$m_{1}g \frac{l_{1}}{2} + m_{2}g \times (1.9) - T \times l_{1}$

$57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (1.9) - T \times 6.5$ = 0

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