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olganol [36]
3 years ago
15

Maria formed when asteroids punctured the moon’s surface, allowing magma to bleed out and create _____.

Physics
2 answers:
vivado [14]3 years ago
5 0

extensive lava flow.


alekssr [168]3 years ago
4 0
Extensive lava flows
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Air enters a turbine operating at steady state at 8 bar, 1600 K and expands to 0.8 bar. The turbine is well insulated, and kinet
kobusy [5.1K]

Answer:

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

Explanation:

To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables

Mathematically this can be determined as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}

Where

Temperature at inlet of turbine

Temperature at exit of turbine

Pressure at exit of turbine

Pressure at exit of turbine

The steady flow Energy equation for an open system is given as follows:

m_i = m_0 = mm(h_i+\frac{V_i^2}{2}+gZ_i)+Q = m(h_0+\frac{V_0^2}{2}+gZ_0)+W

Where,

m = mass

m(i) = mass at inlet

m(o)= Mass at outlet

h(i)= Enthalpy at inlet

h(o)= Enthalpy at outlet

W = Work done

Q = Heat transferred

v(i) = Velocity at inlet

v(o)= Velocity at outlet

Z(i)= Height at inlet

Z(o)= Height at outlet

For the insulated system with neglecting kinetic and potential energy effects

h_i = h_0 + WW = h_i -h_0

Using the relation T-P we can find the final temperature:

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}\\

\frac{T_2}{1600K} = (\frac{0.8bar}{8nar})^{(\frac{1.4-1}{1.4})}\\ = 828.716K

From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK

W = h_i -h_0W = C_p (T_1-T_2)W = 1.005(1600 - 828.716)W = 775.140kJ/Kg

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

4 0
3 years ago
Suppose we have two planets with the same mass, but the radius of the second one is twice the size of the first one. How does th
bogdanovich [222]

The free-fall acceleration on the second planet is one-fourth the value of the first planet.

Calculation:

Consider the mass of planet A to be, M

               the mass of planet B to be, Mₓ = M

               the radius of planet A to be, R₁

               the radius of planet B to be, R₂

The acceleration due to gravity on planet A's surface is given as:

g = GM/R₁²      - (1)

Similarly, the acceleration due to gravity on planet B's surface is given as:

g' = GM/R₂²                           [where, R₂ = 2R₁]

   = GM/4R₁²    -(2)

From equation 1 & 2, we get:

g/g' = GM/R₁² ÷ GM/4R₁²

g/g' = 4/1

Thus we get,

g' = 1/4 g

Therefore, the free-fall acceleration on the second planet is one-fourth the value of the first planet.

Learn more about free-fall here:

<u>brainly.com/question/13299152</u>

#SPJ4

6 0
2 years ago
For a closed system, entropy Group of answer choices may be produced within the system. all of the answers are correct. may be t
wolverine [178]

Answer:

The answer is: all of the answers are correct.

Explanation:

Entropy is defined as the magnitude of the irreversibilities of a process. Entropy is the disorder of molecules within a system. If the system is closed, entropy is produced if the process is irreversible. If the system is subjected to expansions or energy transfer, the entropy increases.

A process is reversible if the system and its surroundings are returned to their initial states. In a quasi-static process it is characterized by being a reversible process in which there are no irreversibilities. The entropy in this system is constant in the system.

6 0
3 years ago
A bartender slides a beer mug at 1.6 m/s towards a customer at the end of a frictionless bar that is 1.1 m tall. The customer ma
Drupady [299]

Answer:

(a): the mug hits the floor 0.752m away from the end of the bar.

(b): the speed of the mug at impact are:

V= 4.87 m/s

direction= 70.82º below the horizontal.

Explanation:

Vx= 1.6 m/s

Vy=?

h= 1.1 m

g= 9.8 m/s²

t is the fall time

t=\sqrt{\frac{2*h}{g} }

t=0.47 sec

Vy= g*t

Vy= 4.6 m/s

V=\sqrt{Vx^{2} +Vy^{2}

V= 4.87 m/s

α= tan⁻¹(Vy/Vx)

α= -70.82º

6 0
2 years ago
Read 2 more answers
A box weighing 18 N requires a force of 6.0 N to drag it at a constant rate. What is the coefficient of sliding friction?
Rzqust [24]
0.33 . Equation is Force of friction equals normal force times coefficient of friction, so 6=18u. Divide 6 by 18
6 0
3 years ago
Read 2 more answers
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