100 J of work in 10 seconds

An ideal gas Carnot cycle with air in a piston cylinder has a high temperature of 1000 K and a heat rejection at 400 K. During t

vivado [14]

Answer:

W / n = - 9133 J / mol, W / n = 3653 J / mol , e = 0.600

Explanation:

The Carnot cycle is described by

$e= 1 - Q_{c} / Q_{H} = 1 - T_{c} / T_{H}$

In this case they indicate that the final volume is

V = 3V₀

In the part of the heat absorption cycle from the source is an isothermal expansion

W = n RT ln (V₀ / V)

W / n = 8.314 1000 ln (1/3)

W / n = - 9133 J / mol

During the part of the isothermal compression in contact with the cold focus, as in a machine the relation of volumes is maintained in this part is compressed three times

W / n = 8.314 400 (3)

W / n = 3653 J / mol

The efficiency of the cycle is

e = 1- 400/1000

e = 0.600

6 0

3 years ago

What type of radiation does not have to be observed above earths atmosphere? Visible light, X rays, Gamma rays, Ultraviolet radi

juin [17]

The correct answer is visible light.

Hope this helps.

6 0

3 years ago

Calculate earth's mass given the acceleration due to gravity at the north pole is 9.830 m/s2 and the mean radius of the earth at

valentinak56 [21]

Answer: $M = 5.98\times 10^{24} kg$

Explanation:

We know that force acting on an object due to Earth's gravity on the surface is given by:

$mg = G\frac{Mm}{r^2}\\ \Rightarrow g = \frac{GM}{r^2}$

where g is the acceleration due to gravity, r would be radius of Earth, M is the mass of Earth and G is the gravitational constant.

It is given that at pole, g = 9.830 m/s² and r = 6371 km = 6371 × 10³ m

$\Rightarrow M = \frac{g\times r^2}{G}$

$M = \frac{9.830 m/s^2 \times (6371 \times 10^3 m )^2}{6.67 \times 10^{-11} m^3 kg^{-1} s^{-2}}$

$\Rightarrow M = 5.98\times 10^{24} kg$

Hence, Earth's mass is $5.98\times 10^{24} kg$

5 0

3 years ago

Structures on a bird feather act like a diffraction grating having 8000 8000 lines per centimeter. What is the angle of the firs

sammy [17]

Answer:

Angle of first order maximum, $\theta=21.19^{\circ}$

Explanation:

Given that,

Wavelength of the light, $\lambda=452\ nm=452\times 10^{-9}\ m$

Number of lines, N = 8000 per cm

The relation between the number of lines and the slit width is given by :

$d=\dfrac{1}{N}$

$d=\dfrac{1}{8000/cm}$

$d=0.000125\ cm=1.25\times 10^{-6}\ m$

The equation of grating is given by :

$d\ sin\theta=n\lambda$

n = 1

$sin\theta=\dfrac{\lambda}{d}$

$sin\theta=\dfrac{452\times 10^{-9}}{1.25\times 10^{-6}}$

$\theta=21.19^{\circ}$

So, the angle of the first-order maximum is 21.19 degrees. Hence, this is the required solution.

6 0

3 years ago

You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfo

dolphi86 [110]

Answer:

Explanation:

To convert gram / centimeter³ to kg / m³

gram / centimeter³

= 10⁻³ kg / centimeter³

= 10⁻³ / (10⁻²)³ kg / m³

= 10⁻³ / 10⁻⁶ kg / m³

= 10⁻³⁺⁶ kg / m³

= 10³ kg / m³

So we shall have to multiply be 10³ with amount in gm / cm³ to convert it into kg/m³

2.33 gram / cm³

= 2.33 x 10³ kg / m³ .

3 0

3 years ago