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MrRa [10]
3 years ago
10

izes water on the surface of Mars? a. Billions of years ago there were lakes and rivers on the surface of Mars, but none exist t

oday. b. There is no evidence that water ever existed on the surface of Mars. c. There have been lakes and ri
Physics
1 answer:
Tanzania [10]3 years ago
6 0

Answer:

The answer is a.

Explanation: Billions of years ago, there were lakes and rivers in the surface of mars because mars had a denser atmosphere and higher surface temperatures which allowed a vast amount of liquid water on the surface. Recently, most of the water on mars today exist as ice.

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What are the musical instrument of liturgy and devitional music.<br><br><br>help me plssss​
kaheart [24]

Answer:  A hymn which accompanies religious observances and rituals.   And others like guitar, violin, flute, harp and the organ

Explanation: Gradually people instructed the natives not only in singing but also in playing various instruments like guitar, violin, flute, harp and later on, the organ.

5 0
2 years ago
What do the three variables (f,m, a) in the equation mean?
Anna [14]

Answer:

f=force m=mass and a=acceleration

7 0
2 years ago
How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe
Rus_ich [418]

Explanation:

(a)  For an isothermal process, work done is represented as follows.

             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

             = -12280.82 \times ln (0.09)

             = -12280.82 \times -2.41

             = 29596.78 J

or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

              = 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c)   We know that for an isothermal process,

               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})

                    = 11 atm

Hence, the required pressure is 11 atm.

(d)   For adiabatic process,  

          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0
3 years ago
When temperature and pressure are applied to sedimentary and igneous rock?
spin [16.1K]
Answer:

Igneous rock

Explanation:

Igneous rocks are formed through the cooling and solidification of magma. It undergoes changes in temperature and pressure that causes it to cool, solidify, and crystallize.
6 0
3 years ago
On your first trip to Planet X you happen to take along a 280 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You'r
arsen [322]

Answer:

5.31143691523 m/s²

Explanation:

m = Mass = 280 g

x = Displacement of spring = 21.7 cm

Time period

T=\dfrac{14}{11}\\\Rightarrow T=1.27\ s

Angular velocity is given by

\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\dfrac{2\pi}{1.27}\\\Rightarrow \omega=4.94739\ rad/s

\omega=\sqrt{\dfrac{k}{m}}\\\Rightarrow k=\omega^2m\\\Rightarrow k=4.94739^2\times 0.28\\\Rightarrow k=6.85346698739\ N/m

From Hooke's law

mg=kx\\\Rightarrow g=\dfrac{kx}{m}\\\Rightarrow g=\dfrac{6.85346698739\times 0.217}{0.28}\\\Rightarrow g=5.31143691523\ m/s^2

The acceleration due to gravity on the planet is 5.31143691523 m/s²

Yes, I have been able to satisfy my curiosity.

7 0
2 years ago
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