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2 years ago

How does energy from the sun affect the motion of molecules in a gas compared to molecules in a liquid?

Physics

2 answers:

Nookie1986 [14]2 years ago

6 0

<h2>Answer:</h2>

Correct option is A.

a. Molecules in a gas move faster than in a liquid.

<h2>Explanation:</h2>

Energy from the sun affect the motion of molecules in a gas compared to molecules in a liquid such as that molecules in a gas move faster than in a liquid. So, first option is correct.

AnnZ [28]2 years ago

3 0

Molecules in a gas move faster than in a liquid.

hope it helps.

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Two long conducting cylindrical shells are coaxial and have radii of 20 mm and 80 mm. The electric potential of the inner conduc

xxMikexx [17]

Answer: 14.52*10^6 m/s

Explanation: In order to explain this problem we have to consider the energy conservation for the electron within the coaxial cylidrical wire.

the change in potential energy for the electron; e*ΔV is equal to energy kinetic gained for the electron so:

e*ΔV=1/2*m*v^2 v^=(2*e*ΔV/m)^1/2= (2*1.6*10^-19*600/9.1*10^-31)^1/2=14.52 *10^6 m/s

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2 years ago

A large crate sits on the floor of a warehouse. Paul and Bob apply constant horizontal forces to the crate. The force applied by

Delicious77 [7]

Answer:

W = -510.98J

Explanation:

Force = 43N, 61° SW

Displacement = 12m, 22° NE

Work done is given as:

W = F*d*cosA

where A = angle between force and displacement.

Angle between force and displacement, A = 61 + 90 + 22 = 172°

W = 43 * 12 * cos172

W = -510.98J

The negative sign shows that the work done is in the opposite direction of the force applied to it.

6 0

2 years ago

Imagine you derive the following expression by analyzing the physics of a particular system: M= (mv2r)(mGr2). Simplify the expre

alex41 [277]

Answer:

The simplified expression is $M = \frac{v^2 r}{G}$

Explanation:

From the question we are told that

$M = \frac{ \frac{m v^2}{r} }{\frac{ mG}{r^2 } }$

So simplifying we have

$M = \frac{m v^2}{r} * \frac{r^2 }{ mG }$

$M = \frac{v^2 r}{G}$

Thus the simplified formula is $M = \frac{v^2 r}{G}$

3 0

2 years ago

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0

2 years ago

Steam undergoes an adiabatic expansion in a piston–cylinder assembly from 100 bar, 360°C to 1 bar, 160°C. What is work in kJ per

vfiekz [6]

Answer:

work is 130.5 kJ/kg

entropy change is 1.655 kJ/kg-k

maximum theoretical work is 689.4 kJ/kg

Explanation:

piston cylinder assembly

100 bar, 360°C to 1 bar, 160°C

to find out

work and amount of entropy and magnitude

solution

first we calculate work i.e heat transfer - work = specific internal energy @1 bar, 160°C - specific internal energy @ 100 bar, 360°C .................1

so first we get some value from steam table with the help of 100 bar @360°C and 1 bar @ 160°C

specific volume = 0.0233 m³/kg

specific enthalpy = 2961 kJ/kg

specific internal energy = 2728 kJ/kg

specific entropy = 6.004 kJ/kg-k

and respectively

specific volume = 1.9838 m³/kg

specific enthalpy = 2795.8 kJ/kg

specific internal energy = 2597.5 kJ/kg

specific entropy = 7.659 kJ/kg-k

now from equation 1 we know heat transfer q = 0

so - w = specific internal energy @1 bar, 160°C - specific internal energy @ 100 bar, 360°C

work = 2728 - 2597.5

work is 130.5 kJ/kg

and entropy change formula is i.e.

entropy change = specific entropy ( 100 bar @360°C) - specific entropy ( 1 bar @160°C )

put these value we get

entropy change = 7.659 - 6.004

entropy change is 1.655 kJ/kg-k

and we know maximum theoretical work = isentropic work

from steam table we know specific internal energy is 2038.3 kJ/kg

maximum theoretical work = specific internal energy - 2038.3

maximum theoretical work = 2728 - 2038.3

maximum theoretical work is 689.4 kJ/kg

3 0

2 years ago