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taurus [48]
3 years ago
9

The nearest star to the Earth is the red dwarf star Proxima Centauri, at a distance of 4.218 light-years. Convert this distance

to barleycorns. (1barleycorn=1/3 inch.)
Physics
1 answer:
alukav5142 [94]3 years ago
3 0

Explanation:

The nearest star to the Earth is the red dwarf star Proxima Centauri, at a distance of 4.218 light-years.  

Light year is the unit of distance covered by the heavenly bodies. 1 light year is equal to :

1\ light\ year=3.72\times 10^{17}\ inches

So, 4.218\ light\ year=4.218\times 3.72\times 10^{17}\ inches

4.218\ light\ year=1.56\times 10^{18}\ inches

We need to convert 4.218 light-years barley corns.  

Since, 1 barleycorn = 1/3 inch  

1\ inch=3\ barleycorn

1.56\times 10^{18}\ inches=3\times 1.56\times 10^{18}=4.68\times 10^{18}\ barleycorn

So, the nearest star to the Earth is at a distance of 4.68\times 10^{18}\ barleycorn. Hence, this is the required solution.

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A skier is pulled by a towrope up a frictionless ski slope that makes an angle of 12 degrees with the horizontal. The rope moves
MArishka [77]

Answer:

Explanation:

Given,

  • Work done by the rope 900 m/s.
  • Angle of inclination of the slope = \theta\ =\ 12^o
  • Initial speed of the skier = v = 1.0 m/s
  • Length of the inclined surface = d = 8.0 m

part (a)

The rope is doing the work against the gravity on the skier to uplift up to the inclined surface. Therefore the work done by the rope is equal to the work done on the skier due to the gravity

\therefore W_r\ =\ W_g\ =\ 900\ J

In both cases the height attained by the skier is equal. and the work done by gravity does not depend upon the speed of the skier.

part (b)

  • Initial speed of the skier = v = 1.0 m/s.

Rate of the work done by the rope is power of the rope.

Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 1.0}{8.0}\\\Rightarrow P\ =\ 112.5\ Watt

Part (c)

  • Initial speed of the skier = v = 2.0 m/s.

Rate of the work done by the rope is power of the rope.

Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 2.0}{8.0}\\\Rightarrow P\ =\ 225\ Watt

4 0
3 years ago
En un balde de 25 cm de altura se hace un orificio a 15 cm del suelo.calcula la velocidad de salida del agua​
Contact [7]

Explanation:

En un balde de 25 cm de altura se hace un orificio a 15 cm del suelo.calcula la velocidad de salida del agua

5 0
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A 2.5-L tank initially is empty, and we want to fill it with 10 g of ammonia. The ammonia comes from a line with saturated vapor
Alex17521 [72]

Answer:

592.92 x 10³ Pa

Explanation:

Mole of ammonia required = 10 g / 17 =0 .588 moles

We shall have to find pressure of .588 moles of ammonia at 30 degree having volume of 2.5 x 10⁻³ m³. We can calculate it as follows .

From the relation

PV = nRT

P x 2.5 x 10⁻³ =  .588 x 8.32 x ( 273 + 30 )

P = 592.92 x 10³ Pa

3 0
3 years ago
The pressure in a traveling sound wave is given by the equation ΔP = (1.78 Pa) sin [ (0.888 m-1)x - (500 s-1)t] Find (a) the pre
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Answer:

a) P_m=1.78\ Pa

b) f=79.5775\ Hz

c) \lambda=7.076\ m

d) v=563.06\ m.s^{-1}

Explanation:

<u>Given equation of pressure variation:</u>

\Delta P= (1.78\ Pa)\ sin\ [(0.888\ m^{-1})x-(500\ s^{-1})t]

We have the standard equation of periodic oscillations:

\Delta P=P_m\ sin\ (kx-\omega.t)

<em>By comparing, we deduce:</em>

(a)

amplitude:

P_m=1.78\ Pa

(b)

angular frequency:

\omega=2\pi.f

2\pi.f=500

∴Frequency of oscillations:

f=\frac{500}{2\pi}

f=79.5775\ Hz

(c)

wavelength is given by:

\lambda=\frac{2\pi}{k}

\lambda=\frac{2\pi}{0.888}

\lambda=7.076\ m

(d)

Speed of the wave is gives by:

v=\frac{\omega}{k}

v=\frac{500}{0.888}

v=563.06\ m.s^{-1}

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