Question #4

If we add 50 Joules of thermal energy to a heat engine, and that heat engine does 30 Joules of work, how much thermal energy is

Natalka [10]

Answer:

The correct answer should be

A. 20 Joules

Explanation:

I'm taking the K12 Unit Test: Energy - Part 1 right now

7 0

3 years ago

Why is temperature scalar?

shepuryov [24]

Temperature is just a measure of how HOT or COLD a substance is, which can be easily defined by a magnitude using a numerical value say “300 K” or “27°C”. Hence we can say it is a scalar quantity.

But the energy which transfer by virtue of a temperature difference is a vector quantity, as it has both magnitude and direction of motion (from High temperature to low temperature region).

3 0

3 years ago

Consider a spherical volume of space that is large enough to be considered homogeneous. Also consider a particle on the surface

LenKa [72]

Answer:

Option A applies.

A. Greater than its escape speed from the mass within the volume

Explanation:

Here it is mentioned that the spherical volume is large enough for the space to be considered as homogeneous. Also, the pressure within the volume is negligible, so that will not result into the re collapse of the Universe. Now as per our knowing, Hubble's Law relates the average speed of the particle to the distance R between the Earth and the particle. So, if the particle's speed is greater than it's escape speed from the mass within the volume, then the Universe is bound to re collapse back again. Option A applies.

3 0

3 years ago

the density of a steel is 7.8.find the mass of steel cube of side 10 cm. find volume of steel if the mass is 8kg

lukranit [14]

Answer:

Mass of the steel cube = 7800 kg

Volume of the steel = 1.025 cubic centimetre

Explanation:

Given:

The density of the steel = 7.8

Side of the cube = 12 cm

<u>(1)The mass of steel cube :</u>

We know that,

$Density = \frac{Mass}{Volume}$

We are given with density and sides of the cube

then volume of the cube

= $(side)^3$

= $10^3$

= 1000 cubic centimetre

Now

$7.8 = \frac{mass}{1000}$

$mass = 7.8 \times 1000$

mass = 7800 kg

<u>(2)volume of steel:</u>

Given the mass = 8 kg

$Density = \frac{Mass}{Volume}$

Substituting the values

$7.8 = \frac{8}{volume}$

$volume = \frac{8}{7.8}$

volume = 1.025 cubic centimetre

3 0

3 years ago

The normal eye, myopic eye and old age

yanalaym [24]

Answer:

1) f’₀ / f = 1.10, the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) the two diameters have the same order of magnitude and are very close to each other

Explanation:

You have some problems in the writing of your exercise, we will try to answer.

1) The equation to be used in geometric optics is the constructor equation

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where p and q are the distance to the object and the image, respectively, f is the focal length

* For the normal eye and with presbyopia

the object is at infinity (p = inf) and the image is on the retina (q = 15 mm = 1.5 cm)

$\frac{1}{f'_o} = 1/ inf + \frac{1}{1.5}$

f'₀ = 1.5 cm

this is the focal length for this type of eye

* Eye with myopia

the distance to the object is p = 15 cm the distance to the image that is on the retina is q = 1.5 cm

1 / f = 1/15 + 1 / 1.5

1 / f = 0.733

f = 1.36 cm

this is the focal length for the myopic eye.

In general, the two focal lengths are related

f’₀ / f = 1.5 / 1.36

f’₀ / f = 1.10

The question of the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) For this second part we have a diffraction problem, the point diameter corresponds to the first zero of the diffraction pattern that is given by the expression for a linear slit

a sin θ= m λ

the first zero occurs for m = 1, as the angles are very small

tan θ = y / f = sin θ / cos θ

for some very small the cosine is 1

sin θ = y / f

where f is the distance of the lens (eye)

y / f = lam / a

in the case of the eye we have a circular slit, therefore the system must be solved in polar coordinates, giving a numerical factor

y / f = 1.22 λ / D

y = 1.22 λ f / D

where D is the diameter of the eye

D = 2R₀

D = 2 0.1

D = 0.2 cm

the eye has its highest sensitivity for lam = 550 10⁻⁹ m (green light), let's use this wavelength for the calculation

* normal eye

the focal length of the normal eye can be accommodated to give a focus on the immobile retian, so let's use the constructor equation

\frac{1}{f} = \frac{1}{p} + \frac{1}{q}

sustitute

$\frac{1}{f} = \frac{1}{25} + \frac{1}{1.5}$

$\frac{1}{f}$ = 0.7066

f = 1.415 cm

therefore the diffraction is

y = 1.22 550 10⁻⁹ 1.415 / 0.2

y = 4.75 10⁻⁶ m

this is the radius, the diffraction diameter is

d = 2y

d_normal = 9.49 10⁻⁶ m

* myopic eye

In the statement they indicate that the distance to the object is p = 15 cm, the retina is at the same distance, it does not move, q = 1.5 cm

$\frac{1}{f} = \frac{1}{15} + \frac{1}{ 1.5}$

$\frac{1}{f}$ = 0.733

f = 1.36 cm

diffraction is

y = 1.22 550 10-9 1.36 10-2 / 0.2 10--2

y = 4.56 10-6 m

the diffraction diameter is

d_myope = 2y

d_myope = 9.16 10-6 m

$\frac{d_{normal}}{d_{myope}}$ = 9.49 /9.16

\frac{d_{normal}}{d_{myope}} = 1.04

we can see that the two diameters have the same order of magnitude and are very close to each other

8 0

3 years ago