Why is it harder to breathe on a

The mean diameters of Mars and Earth are 6.9 ✕ 103 km and 1.3 ✕ 104 km, respectively. The mass of Mars is 0.11 times Earth's mas

Roman55 [17]

Answer:

(a) Ratio of mean density is 0.735

(b) Value of g on mars 0.920 $m,/sec^2$

(c) Escape velocity on earth is $3.563\times 10^4m/sec$

Explanation:

We have given radius of mars $R_{mars}=6.9\times 10^3km=6.9\times 10^6m$ and radius of earth $R_{E}=1.3\times 10^4km=1.3\times 10^7m$

Mass of earth $M_E=5.972\times 10^{24}kg$

So mass of mars $M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg$

Volume of mars $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (6.9\times 10^6)^3=1375.357\times 10^{18}m^3$

So density of mars $d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3$

Volume of earth $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3$

So density of earth $d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3$

(A) So the ratio of mean density $\frac{d_{mars}}{d_E}=\frac{477.69}{649.27}=0.735$

(B) Value of g on mars

g is given by $g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times0.657\times 10^{24}}{(6.9\times 10^6)^2}=0.920m/sec^2$

(c) Escape velocity is given by

$v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec$

5 0

3 years ago

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From a hot air balloon that is at rest at a certain height, a projectile is launched horizontally at 30m / s, how fast will it h

Yuki888 [10]

Answer:

A. 50 m/s

Explanation:

Given in the y direction:

v₀ = 0 m/s

a = 10 m/s²

t = 4 s

Find: v

v = at + v₀

v = (10 m/s²) (4 s) + 0 m/s

v = 40 m/s

In the x direction, the velocity is constant at 30 m/s.

The overall speed is:

v² = (30 m/s)² + (40 m/s)²

v = 50 m/s

3 0

2 years ago

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The process of burning fuel is called?

user100 [1]

The process of burning fuel is Combustion

5 0

3 years ago

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Explain the origin of the magnitude designation for determining the brightness of stars. Why does it seem to go backward, with s

Mashcka [7]

Answer:

Hipparchus was an ancient Greek who classified stars based on the brightness in 129 B.C. He grouped the brightest stars and ranked them 1 (first magnitude) and dimmest stars as 6 (sixth magnitude). Thus, the smaller numbers indicated brighter stars. Now, the scale extends in negative axis as well. More the negative number, brighter is the star. For example, Sun has magnitude -26.74.

This the apparent magnitude which means the classification is based on the brightness of the star as it appears from the Earth.

5 0

3 years ago

A satellite omass1000 kg moves in a circular orbit of radius 8000 km round the earth,assumed to be a sphere of radius 6400 km. C

lubasha [3.4K]

Answer:

ΔE = 37.8 x 10^9 J

Explanation:

The energy required will increased the potential energy and increase the kinetic energy.

As the altitude change is fairly small compared to the earth radius, we can ASSUME that the average gravity will be a good representative

Gravity acceleration at altitude would be 9.8(6400²/8000²) = 6.272 m/s²

G(avg) = (9.8 + 6.272)/2 = 8.036 m/s²

ΔPE = mG(avg)Δh = 1000(8.036)(8e6 - 6.4e6) = 12.857e9 J

The centripetal force at orbit must be equal to the gravity force

mv²/R = mg'

v²/8.0e6 = 6.272

v² = (6.272(8.0e6)) = 50.2e6 m²/s²

The maximum velocity when resting on earth at the equator is about 460 m/s.

The change in kinetic energy is

ΔKE = ½m(vf² - vi²)(1000)

ΔKE = ½(1000)(50.2e6 - 460²) = 25e9 J

Total energy increase is

25e9 + 12.857e9 = 37.8e9 J

3 0

2 years ago