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melomori [17]
2 years ago
7

Consider the first major feature (orderly motions). Which of the following correctly describe patterns of motion in our solar sy

stem? Select all that apply.
(a) The Sun and most of the planets rotate in the same direction that the planets orbit.
(b)Major moons generally have polar orbits, meaning orbits that take them over the north and south poles of the planet they orbit.
(c) Planets have nearly circular orbits.
(d)The outer planets are so large that they nearly collide with each other on each orbit.
(e)All the planets orbit the Sun in nearly the same plane.
(f) Inner planets orbit the Sun in the opposite direction from the outer planets
Physics
1 answer:
snow_lady [41]2 years ago
7 0

Answer:

(a), (c), (e)  

Explanation:

<em>(a) The Sun and most of the planets rotate in the same direction that the planets orbit.</em>

The Sun and most of the planets rotates in the same direction (counter-clockwise) as a consequence of the conservation of the angular momentum from the could in which they were formed.

However, Venus and Uranus rotate in retrograde motion (clockwise), it is  thought that is due to the collision with an asteroid in the early days of both planets.

<em>(b) Major moons generally have polar orbits, meaning orbits that take them over the north and south poles of the planet they orbit.</em>

Polar orbits are not common between major moons, They follow orbits related to the plane of the Solar System.    

<em>(c) Planets have nearly circular orbits.</em>

Kepler's first law establishes:

<em>All the planets revolve around the Sun in an ellipse orbit, with the Sun in one of the focus.</em>

The ellipse has a low eccentricity in the case of planetary orbits.

<em>(d) The outer planets are so large that they nearly collide with each other on each orbit.</em>

They have enough space between them, so it is not possible a collision.

For example, lets take the example of the diameter on Jupiter and Saturn. The diameter of Jupiter is 139.820 Km while Saturn has a diameter of 116.460 Km.

The distance of Saturn from the Sun is 9.5 AU (AU = Astronomical Unit) while in Jupiter is 5.1 AU.

1 AU is defined as the distance between the Earth and the Sun (1.50x10^{8} Km)

The distance of Saturn from the Sun in Kilometers:  

d_{saturn} = 9.5 AU . \frac{1.50x10^{8} Km}{1AU}

d_{saturn} = 1.425x10^{9} Km

The distance of Jupiter from the Sun in Kilometers:

d_{saturn} = 5.1 AU . \frac{1.50x10^{8} Km}{1AU}

d_{saturn} = 7.65x10^{8} Km

Distance between the two planets:

d_{J-S} = 1.425x10^{9} Km -7.65x10^{8} Km

d_{J-S} = 6.60x10^{8} Km

If the distance between the planets is compared with the diameters of any of them, it is easy to see that the planets can't be in a collision since they don't fill the space between them with its diameter.

In the case of Jupiter:

diameter_{Jupiter} d_{J-S}

139.820 Km 6.60x10^{8}

In the case of Saturn:

diameter_{Saturn} d_{J-S}      

116.460 Km 6.60x10^{8}    

<em>(e)All the planets orbit the Sun in nearly the same plane.</em>

When the cloud, from were the planets form, collapse, it starts to rotate and accrete the material in a disk. Given as a result that the planets were in nearly the same plane.

It is important to consider that planets have different degrees of orbital inclinations, which is due to interactions with others object while its formation.

<em>(f) Inner planets orbit the Sun in the opposite direction from the outer planets </em>

This is not correct. Inner planets are constituted by Mercury, Venus, Earth, and Mars.

Mercury, Earth and Mars share the counter-clockwise direction as some outer planets, while Venus shares the same retrograde motion as Uranus.    

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Answer:

The gas was Hexane

Explanation:

taking the diference between the mass of the flask and the final mass qe can calculate the mass of liquid injected (assuming none escaped the flask):

m_{l}  = 27.4593g - 27.0928g = 0.3665g

with the volume of the flask we can get the density of the gas at the indicated pressure and temperature:

d_{g}  = \frac{0.3665 g}{0.1040L} = 3.524 g/L

From the ideal gases law we have that the density can be calculated as:

d_{g}  = \frac{P*M}{R*T}

Where R is the ideal gases constant = , and M the molecular weight of the fluid. Solving for M:

M=\frac{d_{g}*R*T}{P}=\frac{3.524g/L*0,082atmL/molK*291K}{0.976atm}

M=86.16 g/mol

Note that the temperature is computed in Kelvin T= 18+273=291K

The gas with the closer molar mass is Hexane

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2 years ago
4. Dr. Copus is in charge of the cognition department at the University of Wisconsin-Madison. A new drug named Mem-Reen has beco
MrMuchimi

Answer:

See the answer below

Explanation:

<u>Independent variable</u>: Type of drug (Mem-Reen or placebo)

<u>Dependent variable</u>: memories

<u>Experimental group</u>: The group that was given Mem-Reen

<u>Control group</u>: The group that was given placebo

<u>Constants</u>: Food, hours of sleep, memory test procedures.

The independent variable is an input variable that produces effects on the dependent variable. As the variable is changed, it produces different effects on the dependent variable.

The dependent variable is the actual variable that is measured during an experiment. It is the main purpose of setting-up of an experiment.

The experimental group is also referred to as the treatment group while the control group is the group that does not receive treatment at all or they receive fake treatment/placebo.

Constants are unchanging variables included in experiments. They remain unchanged both in the treatment and the control group, otherwise, the outcome of the experiment will be unreliable.

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2 years ago
A student states that 203.9 g of table sugar will form a saturated solution in 100 g of water.
riadik2000 [5.3K]

The temperature and the solubility of sugar at that temperature

Explanation:

The amount of substance which can be dissolved in the solvent depends on the temperature.

As the temperature increases, more substance can be dissolved.

A solution is  saturated if any more of the solute cannot be dissolved in the solution at the given temperature

Hence we need to know the temperature  and also the amount of substance which can be dissolved(solubility) at the same temperature

a) the statement given in option A is correct

b) molar mass has no correlation with the substance's solubility and hence option b is not correct

c) The percent by volume of the solution is not needed to find if the solution is saturated and hence option c is not correct

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A force of 6.0 N pulls a box 0.40 m along a frictionless plane that is inclined at 36°. What work is being done by the pulling f
lys-0071 [83]

Answer:

Expression of work done is

W = Fd cos\theta

Work done to move the sled is given as 1.94 J

Explanation:

As we know that the formula of work done is given as

W = Fd cos\theta

here we know that

F = 6 N

d = 0.4 m

\theta = 36 degree

so we will have

W = 6 \times 0.4 cos36

W = 1.94 J

7 0
2 years ago
Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t
Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

M=-\frac{s'}{s}

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

M=-\frac{15 cm}{12 cm}=-1.25

D. Real and inverted

The magnification equation can be also rewritten as

M=\frac{y'}{y}

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

y'=My

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

M=\frac{5}{9}=-\frac{s'}{s}

which can be rewritten as

s'=-M s = -\frac{5}{9}s

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

s'=-Ms

and then we can substitute it into the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}

and we can solve for s:

\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0
2 years ago
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