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4 years ago

14

Is there a formula for the 2nd Law of Therodynamics, is so name it?

1 answer:

IrinaK [193]4 years ago

8 0

∆S>_closed integral of dQ/T

There are many equations for different situations of entropy but this is a general one

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What is the energy of a photon with a frequency of 1.7 × 1017 Hz? Planck’s constant is 6.63 × 10–34 J•s.

Karo-lina-s [1.5K]

The energy carried by a single photon is given by
$E=hf$
where h is the Planck's constant and f is the frequency of the photon.

The photon of our exercise has a frequency of $f=1.7 \cdot 10^{17} Hz$ , therefore its energy is
$E=hf=(6.63 \cdot 10^{-34}Js)(1.7 \cdot 10^{17} Hz)=1.1 \cdot 10^{-16} J$

3 0

3 years ago

Read 2 more answers

Calculate the acceleration of the car which moves 36m/s in 8s

Semenov [28]

Answer:

the answer to this question is 288

7 0

3 years ago

The near point of an eye is 110 cm. A corrective lens is to be used to allow this eye to focus clearly on objects 26.0 cm in fro

irga5000 [103]

Answer:

The focal length of the lens is 34.047 cm

The power of the needed corrective lens is 2.937 diopter.

Explanation:

Distance of the object from the lens,u = 26 cm

Distance of the image from the lens ,v= -110 cm

(Image is forming on the other side of the lens)

Since ,lens of the human eye is converging lens,convex lens.

Using a lens formula:

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

$\frac{1}{f}=\frac{1}{26 cm}+\frac{1}{-110 cm}$

f = 34.047 cm = 0.3404 m

Power of the lens = P

$P=\frac{1}{f}=\frac{1}{0.34047 m}=2.937 diopter$

6 0

3 years ago

A racemic mixture can rotate a plane-polarized light in a clockwise direction. Select one: True False

laiz [17]

Answer:

True

Explanation:

8 0

3 years ago

Jupiter's satellite Europa orbits Jupiter with a period of 3.55 d at an orbital radius of 6.71 108 m (assume the orbit to be cir

yawa3891 [41]

Answer:

(a)

M = 1.898 x 10^27 kg

(b)

v = 13.74 km/s

(c) E = 0.28 N/kg

Explanation:

Time period, T = 3.55 days = 3.55 x 24 x 3600 second = 306720 s

Radius, r = 6.71 x 10^8 m

G = 6.67 x 10^-11 Nm^2/kg^2

(a) $T=2\pi \sqrt{\frac{r^{3}}{GM}}$

$M=\frac{4\pi ^{2}r^{3}}{GT^{2}}$

$M=\frac{4\times3.14^{2}\times 6.71^{3}\times 10^{24}}{6.67\times 10^{-11}\times 306720^{2}}$

M = 1.898 x 10^27 kg

(b) Let v be the orbital velocity

$v=\frac{2\pi r}{T}$

$v=\frac{2\times 3.14\times 6.71\times 10^{8}}{306720}$

v = 13739.5 m/s

v = 13.74 km/s

(b) The gravitational field E is given by

$E = \frac{GM}{r^{2}}$

$E = \frac{6.67\times10^{-11}\times 1.898\times 10^{27}}{6.71^{2}\times 10^{16}}$

E = 0.28 N/kg

6 0

3 years ago