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3 years ago

Mathematical expression between wave number and frequency

Physics

1 answer:

pav-90 [236]3 years ago

7 0

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A student stands with both feet on some scales in order to measure his weight.

Neko [114]

Answer:

500 N

Explanation:

Because even if he lift one foot his weight will be same as the pressure applied on the scale will be the same and will not change it is not that the scale measures each foot separately

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3 years ago

A civil engineer wishes to redesign the curved roadway in the example What is the Maximum Speed of the Car? in such a way that a

vlabodo [156]

Answer:

24.3 degrees

Explanation:

A car traveling in circular motion at linear speed v = 12.8 m/s around a circle of radius r = 37 m is subjected to a centripetal acceleration:

$a_c = \frac{v^2}{r} = \frac{12.8^2}{37} = 4.43 m/s^2$

Let α be the banked angle, as α > 0, the outward centripetal acceleration vector is split into 2 components, 1 parallel and the other perpendicular to the road. The one that is parallel has a magnitude of 4.43cosα and is the one that would make the car slip.

Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.

So $gsin\alpha = 4.43cos\alpha$

$\frac{sin\alpha}{cos\alpha} = \frac{4.43}{g} = \frac{4.43}{9.81} = 0.451$

$tan\alpha = 0.451$

$\alpha = tan^{-1}0.451 = 0.424 rad = 0.424*180/\pi \approx 24.3^0$

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4 years ago

A sealed pressure cooker allows steam to escape only through an opening in the lid. a separate metal object, called a petcock, s

daser333 [38]

The mass of the petcock of the pressure cooker is 0.257 kg.

The total pressure inside the cooker is= atmospheric pressure+Gauge pressure

=100 kPa+101 kPa=201*10^3 Pa

The area of the petcock is= (π/4)d^2=0.785*0.004^2=12.56*10^(-6) m^2

Now the mass of the petcock is calculated as

P=(F/A)=(mg/A)

201*10^3=(m*9.81/(12.56*10^(-6)

m=0.257 kg

Therefore the mass of the petcock is 0.257 kg

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4 years ago

A piston contains oxygen at 280 k occupying a volume of 0.25m3. The cylinder is compressed adiabatically to 0.14 m3, find the in

spayn [35]

dhvxdryvbjjolbcsdfunnlpifsaf

8 0

3 years ago

. A long 10-cm-diameter steam pipe whose external surface temperature is 110oC passes through some open area that is not protect

Nata [24]

Answer:

$Nu = 30.311$

Explanation:

Let consider that pipe is a horizontal cylinder. The Nusselt number is equal to:

$Nu = \left\{0.6+\frac{0.387\cdot Ra_{D}^{\frac{1}{6} }}{[1+\left(\frac{0.559}{Pr} \right)^{\frac{9}{16}} ]^{\frac{8}{27} }} \right\}^{2}$ , for $Ra_{D} \le 10^{12}$ .

Where $Ra_{D}$ is the Rayleigh number associated with the cylinder.

The Rayleigh number is:

$Ra_{D} = \frac{g\cdot \beta\cdot (T_{pipe}-T_{air})\cdot D^{3}}{\nu^{2}}\cdot Pr$

By assuming that air behaves ideally, the coefficient of volume expansion is:

$\beta = \frac{1}{T}$

$\beta = \frac{1}{283.15\,K}$

$\beta = 3.532\times 10^{-3}\,\frac{1}{K}$

The cinematic and dynamic viscosities, thermal conductivity and isobaric specific heat of air at 10 °C and 1 atm are:

$\nu = 1.426\times 10^{-5}\,\frac{m^{2}}{s}$

$\mu = 1.778\times 10^{-5}\,\frac{kg}{m\cdot s}$

$k = 0.02439\,\frac{W}{m\cdot ^{\textdegree}C}$

$c_{p} = 1006\,\frac{J}{kg\cdot ^{\textdegree}C}$

The Prandtl number is:

$Pr = \frac{\mu\cdot c_{p}}{k}$

$Pr = \frac{(1.778\times 10^{-5}\,\frac{kg}{m\cdot s} )\cdot (1006\,\frac{J}{kg\cdot ^{\textdegree}C} )}{0.02439\,\frac{W}{m\cdot ^{\textdegree}C} }$

$Pr = 0.733$

Likewise, the Rayleigh number is:

$Ra_{D} = \frac{(9.807\,\frac{m}{s^{2}} )\cdot (3.532\times 10^{-3}\,\frac{1}{K} )\cdot (110^{\textdegree}C-10^{\textdegree}C)\cdot (0.1\,m)^{3}}{(1.426\times 10^{-5}\,\frac{m^{2}}{s})^{2} }\cdot (0.733)$

$Ra_{D} = 12.486\times 10^{6}$

Finally, the Nusselt number is:

$Nu = \left\{0.6+\frac{0.387\cdot (12.486\times 10^{6})^{\frac{1}{6} }}{\left[1 + \left(\frac{0.559}{0.733}\right)^{\frac{9}{16} }\right]^{\frac{8}{27} }} \right\}^{2}$

$Nu = 30.311$

8 0

3 years ago

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Mathematical expression between wave number and frequency​

Mathematical expression between wave number and frequency