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mash [69]
2 years ago
13

The temperature and pressure at the surface of Mars during a Martian spring day were determined to be -41 oC and 900 Pa, respect

ively. (a) Determine the density of the Martian atmosphere for these conditions if the gas constant for the Martian atmosphere is assumed to be equivalent to that of carbon dioxide. (b) Compare the answer from part (a) with the density of the earth’s atmosphere during a spring day when the temperature is 18 oC and the pressure 101.6 kPa (abs).
Physics
1 answer:
Fittoniya [83]2 years ago
8 0

Answer:

Part a)

\rho = 0.0205 kg/m^3

Part b)

\rho = 1.22 kg/m^3

So density of atmosphere at Martian Surface is very less than the density at Earth.

Explanation:

Part a)

As per ideal gas equation we know that

PM = \rho RT

here we know that Martian atmosphere is equivalent to that of carbon

so we will have

P = 900 Pa

T = 273 - 41 = 232 K

now we will have

(900)(0.044) = \rho (8.31)(232)

\rho = 0.0205 kg/m^3

Part b)

Now for the earth surface the density of air is given for

P = 101.6 kPa

T = 18 ^oC

so we will have

PM = \rho RT

(101.6\times 10^3)(0.029) = \rho(8.31)(273 + 18)

\rho = 1.22 kg/m^3

So density of atmosphere at Martian Surface is very less than the density at Earth.

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In a lever, the effort arm is two times as long as the load arm. The resultant force will be
jasenka [17]
Answer is B. 

In a lever, the effort arm is 2 times as a long as the load arm. The resultant force will be twice the applied force.

Hope it helped you.

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Charlie
5 0
3 years ago
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3. A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of
Alex Ar [27]

Answer:

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

Explanation:

Given that,

The respective indices of refraction for the blue light and the red light are 1.4636 and 1.4561.

A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of glass at 80 degrees.

We need to find the angular separation between the refracted red and refracted blue beams while they are in the glass.

Using Snell's law for red light as :

n_1\sin\theta_1=n_2\sin\theta_2\\\\\theta_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta_2=\sin^{-1}((\dfrac{1}{1.4561})\sin(80))\\\\\theta_2=42.555

Again using Snell's law for blue light as :

n_1\sin\theta_1=n_2\sin\theta'_2\\\\\theta'_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta'_2=\sin^{-1}((\dfrac{1}{1.4636 })\sin(80))\\\\\theta'_2=42.283

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

7 0
3 years ago
I need help with #4 and #5?
algol [13]

A

Explanation:

just  pick a

3 0
2 years ago
A particle is released as part of an experiment. Its speed t seconds after release is given by v (t )equalsnegative 0.4 t square
torisob [31]

Answer:

a) 2.933 m

b) 4.534 m

Explanation:

We're given the equation

v(t) = -0.4t² + 2t

If we're to find the distance, then we'd have to integrate the velocity, since integration of velocity gives distance, just as differentiation of distance gives velocity.

See attachment for the calculations

The conclusion of the attachment will be

7.467 - 2.933 and that is 4.534 m

Thus, The distance it travels in the second 2 sec is 4.534 m

6 0
3 years ago
7 The weight of a space shuttle is about 20.0 million newtons. What is the space shuttle's weight when it is traveling through o
kogti [31]

Well first of all, the Space Shuttle program ended a few years ago, and none have been launched since then.

The Shuttle never went to places that were properly referred to as "outer space". When they flew, the Space Shuttles went to low Earth orbit, where the acceleration of gravity is roughly 85% of its value on the Earth's surface.

So a Shuttle that weighed 20 million Newtons on the launch pad weighed roughly  17 million Newtons while in orbit.


4 0
3 years ago
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