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3 years ago

Our molecular model of matter describes a fluid as consisting of lots of little particles (atoms or molecules)

Physics

1 answer:

sergij07 [2.7K]3 years ago

5 0

Answer:

a) μ = [kg / m s]

b) D = m²/s

Explanation:

Viscosity is equivalent to a friction force that opposes the movement of fluids, it is defined by the relation

Shear stress = μ speed gradient

Fr/A = μ v₀ / h

Where Fr is the shear force, A the area, h the height and v₀ the velocity of the fluid layer

μ = [kg / m s]

The diffusion coefficient characterizes the ability of a material to move in a given solvent, depending on the size of the solute, the viscosity of the solvent, temperature

J = - D dΘ / dx

J is the broadcast glare

D = [m² / s]

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Assume a satellite shines an unpolarized light on a telescope. The intensity of the light as it reaches the telescope is 1.1*10-

n200080 [17]

Answer:

$4.125\times 10^{-11}\ W/m^2$

Explanation:

$I_0$ = Intensity of unpolarized light = $1.1\times 10^{-10}\ W/m^2$

$\theta$ = Angle of the filter = $30^{\circ}$

Intensity of light is given by

$I=\dfrac{I_0}{2}cos^2\theta\\\Rightarrow I=\dfrac{1.1\times 10^{-10}}{2}cos^230\\\Rightarrow I=4.125\times 10^{-11}\ W/m^2$

The intensity of light detected by the camera is $4.125\times 10^{-11}\ W/m^2$

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4 years ago

What will be the final temperature if a 4.00 g silver ring at 41.0◦C if it gives off 18.0 J of heat to the surroundings? The spe

Eduardwww [97]

Answer:

Final temperature, $T_f=21.85^{\circ}$

Explanation:

Given that,

Mass of silver ring, m = 4 g

Initial temperature, $T_i=41^{\circ}C$

Heat released, Q = -18 J (as heat is released)

Specific heat capacity of silver, $c=0.235\ J/g\ C$

To find,

Final temperature

Solution,

The expression for the specific heat is given by :

$Q=mc\Delta T$

$Q=mc(T_f-T_i)$

$T_f=\dfrac{-Q}{mc}+T_i$

$T_f=\dfrac{-18}{4\times 0.235}+41$

$T_f=21.85^{\circ}$

So, the final temperature of silver is 21.85 degrees Celsius.

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3 years ago

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VikaD [51]

Answer: it should be c and d

Explanation:

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