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3 years ago

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Given that water at standard pressure freezes at 0∘c, which corresponds to 32∘f, and that it boils at 100∘c, which corresponds t

o 212∘f, calculate the temperature difference δt in degrees fahrenheit that corresponds to a temperature difference of 1 k on the kelvin scale.

1 answer:

Zolol [24]3 years ago

6 0

The temperature difference of 1 K is equivalent to the temperature difference of 1 °C. Therefore, we find the relationship between the change in °F and °C.
A change of 212 - 32 °F is the same as a change of 100 - 0 °C. Thus:
(212 - 32) °F = (100 - 0) °C
1 °C = 1.8 °F
1 K = 1.8 °F

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Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

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The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

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(b) What is the mass of each box?

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(b) m1 = 915kg;

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Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u> $m_{1}$ <u>:</u>

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

<u>For </u> $m_{2}$ <u>:</u>

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

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<u>For </u> $m_{3}$ <u>:</u>

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

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Answer:

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