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3 years ago

A person who has a job that requires working all night and sleeping during the day will end up fighting hos or her natural

Physics

2 answers:

Andreyy893 years ago

7 0

Answer:

The answer is circadian rhythm (A).

LuckyWell [14K]3 years ago

5 0

Answer:

The answer to this question is circadian rhythm. A person who has a job that requires working all night and sleeping during the day will end up fighting his or her natural circadian rythm. Hope it helps!

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A particle moves along the curve below. y = sqrt(1 + x^3) As it reaches the point (2, 3), the y-coordinate is increasing at a ra

blagie [28]

Answer:7 cm/s

Explanation:

Given

Particle move along curve

$y=\sqrt{1+x^3}$

As it reaches the (2,3) its y coordinate is increasing at 14 cm/s

Differentiating y w.r.t time

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3x^2}{2\sqrt{1+x^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

Now at (2,3)

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3\cdot 2^2}{2\sqrt{1+2^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$14=\frac{3\times 4}{2\times \sqrt{9}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{\mathrm{d} x}{\mathrm{d} t}=7 cm/s$

7 0

3 years ago

Which best describes the relationship between the direction of energy and wave motion in a transverse wave?

sammy [17]

I think the correct answer from the choices listed above is the second option. The relationship between the direction of energy and wave motion in a transverse wave would be the energy direction is perpendicular to the motion of the wave. Hope this answers the question. Have a nice day.

6 0

3 years ago

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A JFET has a drain current of 5mA. If IDSS = 10mA and VGS ( off )= -6 v. find The Value Of

levacccp [35]

$\underline {\huge \boxed{ \sf \color{skyblue}Answer : }}$

Given :

$\tt \large {\color{purple} ↬ } \: \: \: \: \: I_{D} = 5mA$

$\: \:$

$\tt \large {\color{purple} ↬ } \: \: \: \: \: I_{DSS} = 10mA$

$\: \:$

$\tt \large {\color{purple} ↬ } \: \: \: \: \: V_{GS(off)} = -6V$

$\: \:$

$\tt \large {\color{purple} ↬ } \: \: \: \: \: V_{GS} = {?}$

$\: \: \:$

Let's Slove :

$\tt \large I_{D} = I_{(DSS)} (1 - \frac {V_{GS}}{V_{GS(off)}} )^{2}$

$\: \: \:$

$\tt \large \: V_{GS} = (1 - \frac{ \sqrt{I_D} }{ \sqrt{I_{DSS}} } ) \times V_{GS(off)}$

$\: \: \:$

$\tt \large \: V_{GS} = (1 - \frac{ \sqrt{5m} }{ \sqrt{10m} } ) \times { - 6}$

$\: \:$

$\underline \color{red} {\tt \large \boxed {\tt V_{GS} = 1.75 ✓}}$

3 0

1 year ago

Help with this review pls!!

o-na [289]

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5 0

3 years ago

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The ratio of the focal length of a lens to it's diameter is the

user100 [1]

Answer:

The f-ratio describes the relationship between the lens diameter and the focal length and is calculated by dividing the focal length by the diameter of the lens. For example, if a lens were to have a focal length of 50mm and a diameter of 10mm, then the f-ratio would be 50mm/10mm=5 or otherwise referred to as f5.

Explanation:

6 0

2 years ago

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