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Vilka [71]
3 years ago
11

Compare the atmospheres of Mars and Venus. Mars is rich in oxygen, like ours, accounting for its red surface. Like Earth, nitrog

en is the chief atmospheric gas. Both are too hot for water to now exist as a liquid at the surface. Both are chiefly carbon dioxide, but at Mars it can freeze as dry ice. Both are made of hydrogen and helium, like the jovians.
Physics
1 answer:
enyata [817]3 years ago
5 0

Answer:

Explanation:

Venus's atmosphere is very thick, dry and hot whereas Mars's atmosphere is very thin and cold.

Both Venus's and Mars's atmospheres are about 95 percent carbon dioxide.

The surface temperature of Venus is around 890 degrees F, the hottest average temperature in the Solar System. This is due to abundance of greenhouse gasses. The atmosphere of Venus is composed of 97% CO2, 2% N2 and less than 1% of O2, H2O and CH4 (methane). Since CO2 is a major greenhouse gas, the radiation from the Sun is trapped in the atmosphere of Venus producing an extremely high surface temperature.

Mars has an atmospheric composition of 95% CO2, 3% N2, 2% Ar and less than 1% O2.A high noble gas content implies that Mar's atmosphere was much thicker in the past (noble gases do not react with other elements and are heavy enough to stay within the gravitational field of Mars). The climate on Mars is very desert-like due to its thin atmosphere. There is too little mass in the atmosphere to hold in heat so the warmest daytime temperatures are around 50 degrees F, but the nighttime temperatures are -170 degrees F. Other weather features are massive dust storms and occasional CO2 fog in the canyons.

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Rutherford and atomic model are correctly matched.

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The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the locat
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E=0

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Which of the following is a potential result of preparing appropriately before starting an experiment in a lab?
oksano4ka [1.4K]

Answer:

Your project goes well.

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3 years ago
A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra
Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

sin(\alpha) = \frac{Vyi}{Vi}

Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}

Vy in this problem will follow this equation =

Vy(t) = Vyi -g.t

where g is the gravity acceleration

Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t

This is equation (1)

For Y(t) :

Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}

We suppose yi = 0

Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s

So in t = 0.675 s  → Vy = 0. Now we calculate the y in which this happen using (2)

Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} }  .(0.675s)^{2} \\Y(0.675s) =2.236 m

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0
3 years ago
What is the equation to find average velocity?
seraphim [82]
The mean of the velocities.
(finalVelocity + initialVelocity)/2
3 0
2 years ago
Read 2 more answers
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