A heat engine is used to convert heat into which of the following?

Answer the following. (a) What is the surface temperature of Betelgeuse, a red giant star in the constellation of Orion, which r

bagirrra123 [75]

Answer:

(a) T = 2987.6 k

(b) T = 19986.2 k

Explanation:

The temperature of a star in terms of peak wavelength can be given by Wein's Displacement Law, which is as follows:

$T = \frac{0.2898\ x\ 10^{-2}\ m.k}{\lambda_{max}}$

where,

T = Radiated surface temperature

$\lambda_{max}$ = peak wavelength

(a)

here,

$\lambda_{max}$ = 970 nm = 9.7 x 10⁻⁷ m

Therefore,

$T = \frac{0.2898\ x\ 10^{-2}\ m.k}{9.7\ x\ 10^{-7}\ m}$

T = 2987.6 k

(b)

here,

$\lambda_{max}$ = 145 nm = 1.45 x 10⁻⁷ m

Therefore,

$T = \frac{0.2898\ x\ 10^{-2}\ m.k}{1.45\ x\ 10^{-7}\ m}$

T = 19986.2 k

6 0

2 years ago

The greater the mass of an object, the greater its force due to gravity

Serga [27]

What is the question?

6 0

2 years ago

At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107

dangina [55]

Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

$K = - V\cdot \frac{dP}{dV}$

Where:

$K$ - Bulk module, measured in pascals.

$V$ - Sample volume, measured in cubic meters.

$P$ - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

$-\frac{K \,dV}{V} = dP$

This resultant expression is solved by definite integration and algebraic handling:

$-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP$

$-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}$

$\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}$

$\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }$

The final volume is predicted by:

$V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }$

If $V_{o} = 1\,m^{3}$ , $P_{o} - P_{f} = -10132500\,Pa$ and $K = 2.3\times 10^{9}\,Pa$ , then:

$V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }$

$V_{f} \approx 0.996\,m^{3}$

Change in volume due to increasure on pressure is:

$\Delta V = V_{o} - V_{f}$

$\Delta V = 1\,m^{3} - 0.996\,m^{3}$

$\Delta V = 0.004\,m^{3}$

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

8 0

3 years ago

What is the current if a charge of 25 coulombs passes in 5 seconds?

Jobisdone [24]

Answer: 5 amps

Explanation: I = Q/t

8 0

2 years ago

The first antiparticle, the positron or antielectron, was discovered in 1932. It had been predicted by Paul Dirac in 1928, thoug

Taya2010 [7]

Answer:

Energy of Photon = 4.091 MeV

Explanation:

From the conservation of energy principle, we know that total energy of the system must remain conserved. So, the energy or particles before collision must be equal to the energy of photons after collision.

K.E OF electron + Rest Energy of electron + K.E of positron + Rest Energy of positron = 2(Energy of Photon)

where,

K.E OF electron = 3.58 MeV

Rest Energy of electron = 0.511 MeV

Rest Energy of positron = 0.511 MeV

K.E OF positron = 3.58 MeV

Energy of Photon = ?

Therefore,

3.58 MeV + 0.511 MeV + 3.58 MeV + 0.511 MeV = 2(Energy of Photon)

Energy of Photon = 8.182 MeV/2

Energy of Photon = 4.091 MeV

8 0

2 years ago