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spayn [35]
2 years ago
9

A patient comes to an outpatient laboratory for a physician-ordered fasting test. The patient indicates that he forgot that the

test was to be fasting and ate a candy bar 2 hours ago. The patient insists that you should draw the test because he cannot come back at another time. What should you do?
Physics
1 answer:
Fynjy0 [20]2 years ago
3 0

The patient eating a candy bar instead of fasting for the test should be told

that the test results will be wrong and he may receive a wrong diagnosis.

The medical practitioners told him to fast when coming for a reason and he

forgetting and eating something means the objective for the test may have

been defeated.

This is why it's best to explain to him about the consequences of his actions

which is likely getting a wrong result and diagnosis.

Rad more on brainly.com/question/19622125

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An engineer is designing a runway. She knows that a plane, starting at rest, needs to reach a speed of 180mph at take-off. If th
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Answer:

The plane would need to travel at least 8,\!580\; {\rm ft} (8.58 \times 10^{3}\; {\rm ft}.)

The 10,\!000\; {\rm ft} runway should be sufficient.

Explanation:

Convert unit of the the take-off velocity of this plane to \rm ft\cdot s^{-1}:

\begin{aligned}v &= 180\; {\rm mph} \\ &= 180\; {\rm mi \cdot hrs^{-1}} \times \frac{1\; {\rm hrs}}{3600\; {\rm s}} \times \frac{5280\; {\rm ft}}{1\; {\rm mi}} \\ &= 264\; {\rm ft \cdot s^{-1}}\end{aligned}.

Initial velocity of the plane: u = 0\; {\rm ft \cdot s^{-1}}.

Take-off velocity of the plane v =264\; {\rm ft\cdot s^{-1}}.

Let x denote the distance that the plane travelled along the runway. Since acceleration is constant but unknown, make use of the SUVAT equation x = ((u + v) / 2) \, t.

Notice that this equation does not require the value of acceleration. Rather, this equation make use of the fact that the distance travelled (under constant acceleration) is equal to duration t times average velocity (u + v) / 2.

The distance that the plane need to cover would be:

\begin{aligned}x &= \left(\frac{u + v}{2}\right)\, t \\ &= \frac{0\; {\rm ft \cdot s^{-1}} + 264\; {\rm ft \cdot s^{-1}}}{2} \times 65.0\; {\rm s} \\ &= 8.58\times 10^{3}\; {\rm ft}\end{aligned}.

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