Hi I’m having some difficulty with this question could really use some help!

Which of the following are true about a "simple compressible system"? It cannot be a mixture of different substances (e.g. oxyge

fomenos

Explanation:

The volume of a simple compressible system is not fixed. At a state of equilibrium, there should be uniformity in the entire system.

From the question we have here, these are the correct options:

1. It cannot be a mixture of different substances (e.g. oxygen and nitrogent)

2. It can be composed of any phases of a substance: solid, liquid, and/or gas

3. It's state is specified if given two independent, intensive thermodynamic properties.

4 0

3 years ago

A paperweight consists of a 8.55-cm-thick plastic cube. Within the plastic a thin sheet of paper is embedded, parallel to opposi

ella [17]

Answer:

$\mu=1.5322$

Explanation:

Given:

Thickness of the paperweight cube, $x=8.55\ cm$

apparent depth from one side of the inbuilt paper in the plastic cube, $i=4.02\ cm$

apparent depth from the other side of the inbuilt paper in the plastic cube, $i'=1.56\ cm$

Now as we know that refractive index is given as:

$\rm \mu=\frac{real\ depth}{apparent\ depth}$

Let the real depth form first side of the slab be, $d$

Then the depth from the second side of the slab will be, $d'=x-d=8.55-d$

Since refractive index for an amorphous solid is an isotropic quantity so it remains same in all the direction for this plastic.

$\mu=\mu'$

$\frac{d}{i} =\frac{d'}{i'}$

$\frac{d}{4.02}=\frac{8.55-d}{1.56}$

$d=6.1597\ cm$

Now the refractive index:

$\mu=\frac{d}{i}$

$\mu=\frac{6.1596}{4.02}$

$\mu=1.5322$

7 0

3 years ago

A steady 45 N horizontal force is applied to a 15 kg object a table. The object slides against a friction force of 30 N. Calcula

lisov135 [29]

The net force on an object subject to friction is equal to the sum of the applied force and the frictional force.

Mathematically,

$F_{N} = ma = F_{applied} - f_{fr}$

Here, m is mass of object and a is its acceleration. We take frictional force negative because it opposes the motion of object.

Given, $m=15\ kg$ , $F_{applied} =45\ N$ and $f_{fr} = 30\ N$

Substituting these values in above formula, we get

$15\ kg\times a = 45\ N -30\ N=15\ N \\\\a=\frac{15\ N}{15\ kg} =1\ m/s^2$ .

Thus, the acceleration of an object is $1\ m/s^2.$

6 0

3 years ago

Is the angular position of the first-order spectrum small enough for sinθ≈θ to be a good approximation?.

vesna_86 [32]

Answer:

15 to 30 so anywhere in between there.

7 0

2 years ago

Two factors effecting the magnitude of the force of gravity between 2 objects are...

slavikrds [6]

Answer:

Two factors effecting the magnitude of the force of gravity between 2 objects are the product of their masses and square of distance between them.

Explanation:

According to Newton's law of universal gravitation

$F = G\frac{m_{1}m_{2} }{r^{2} } }$

where F is the gravitational force, G is the universal gravitational constant and its value is 6.6743 × 10⁻¹¹ Nm²/kg₂ , m₁ and m₂ are masses of bodies and r is the distance between them.

It can be seen from the above equation that F is directly proportional to the product of the masses and inversely proportional to the square of distance between them.

F ∝ m₁m₂

F ∝ 1/r²

As far as the masses of the bodies increase, magnitude of the Gravitational force increases and if distance between them increase then Gravitational force between them decreases.

5 0

3 years ago