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3 years ago

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Mars moves in an elliptical orbit around the sun, and the mass of mars is much less than the mass of the sun. at the instant sho

wn above, mars is getting farther away from the sun. how does this affect the potential energy of the mars-sun system and the magnitude of mars angular momentum with respect to the sun

1 answer:

Bess [88]3 years ago

4 0

I believe the answer would be that the system potential energy will increase; while the Mar's angular momentum will remain the same.
Mars rotates on its axis, completing one revolution every 24.6 hours. The axis of mars is tilted at 25 degrees and 12 minutes relative to its orbital plane about the Sun. This produces seasons on the surfaces of Mars, similar to the seasons on Earth.

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The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold 24 x (mass o

matrenka [14]

Answers:

(a) $0.0073kg$

(b) Volume gold: $3.79(10)^{-7}m^{3}$ , Volume cupper: $7.6(10)^{-8}m^{3}$

(c) $17633.554kg/m^{3}$

Explanation:

<h2>(a) Mass of gold </h2><h2 />

We are told the total mass $M$ of the coin, which is an alloy of gold and copper is:

$M=m_{gold}+m_{copper}=7.988g=0.007988kg$ (1)

Where $m_{gold}$ is the mass of gold and $m_{copper}$ is the mass of copper.

In addition we know it is a 22-karat gold and the relation between the number of karats $K$ and mass is:

$K=24\frac{m_{gold}}{M}$ (2)

Finding ${m_{gold}$ :

$m_{gold}=\frac{22}{24}M$ (3)

$m_{gold}=\frac{22}{24}(0.007988kg)$ (4)

$m_{gold}=0.0073kg$ (5) This is the mass of gold in the coin

<h2>(b) Volume of gold and cupper</h2><h2 />

The density $\rho$ of an object is given by:

$\rho=\frac{mass}{volume}$

If we want to find the volume, this expression changes to: $volume=\frac{mass}{\rho}$

For gold, its volume $V_{gold}$ will be a relation between its mass $m_{gold}$ (found in (5)) and its density $\rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}$ :

$V_{gold}=\frac{m_{gold}}{\rho_{gold}}$ (6)

$V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}$ (7)

$V_{gold}=3.79(10)^{-7}m^{3}$ (8) Volume of gold in the coin

For copper, its volume $V_{copper}$ will be a relation between its mass $m_{copper}$ and its density $\rho_{copper}=8.96g/cm^{3}=8960kg/m^{3}$ :

$V_{copper}=\frac{m_{copper}}{\rho_{copper}}$ (9)

The mass of copper can be found by isolating $m_{copper}$ from (1):

$M=m_{gold}+m_{copper}$

$m_{copper}=M-m_{gold}$ (10)

Knowing the mass of gold found in (5):

$m_{copper}=0.007988kg-0.0073kg=0.000688kg$ (11)

Now we can find the volume of copper:

$V_{copper}=\frac{0.000688kg}{8960kg/m^{3}}$ (12)

$V_{copper}=7.6(10)^{-8}m^{3}$ (13) Volume of copper in the coin

<h2>(c) Density of the sovereign coin</h2><h2 />

Remembering density is a relation between mass and volume, in the case of the coin the density $\rho_{coin$ will be a relation between its total mass $M$ and its total volume $V$ :

$\rho_{coin}=\frac{M}{V}$ (14)

Knowing the total volume of the coin is:

$V=V_{gold}+V_{copper}=3.79(10)^{-7}m^{3}+7.6(10)^{-8}m^{3}=4.53(10)^{-7}m^{3}$ (15)

$\rho_{coin}=\frac{0.007988kg}{4.53(10)^{-7}m^{3}}$ (16)

Finally:

$\rho_{coin}=17633.554kg/m^{3}}$ (17) This is the total density of the British sovereign coin

6 0

3 years ago

which has a higher acceleration:a 10kg object acted upon with a net force of 20N or an 18kg object acted on by a net force of 20

MA_775_DIABLO [31]

<span>Answer: The acceleration of 10 kg object is greater than that of 18 kg object.

Explanation:
According to Newton's Second law:
F = ma --- (A)

Let's find the acceleration for both 10 kg and 18 kg objects!
The net force on both of these masses = F = 20N

(1) Acceleration of 10 kg object
Mass = m = 10 kg
Plug in the values in equation (A):
20 = 10 * a
Acceleration = a = 2 m/s^2

(2) Acceleration of 18 kg object
Mass = m = 18 kg
Plug in the values in equation (A):

20 = 18 * a
Acceleration = a = 1.11 m/s^2

2 > 1.11; therefore, 10 kg object has the higher acceleration compared to the acceleration of the 18 kg object.</span>

7 0

3 years ago

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Someone help with this table!! 50 points!!!! Please help ASAP!!!!!!

Vesnalui [34]

I have the same thing for homework

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3 years ago

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Before using wind energy to produce electricity, humans used wind energy to produce PLEASE ANSWER QUICKLY

topjm [15]

Answer:

Water

Explanation:

windmills were used to bring water to farms in ancient times.

8 0

3 years ago

If earth increase the distance from the sun, what will happen to the period of orbi t(the time it takes to complete one revoluti

Mandarinka [93]

The period of the orbit would increase as well

Explanation:

We can answer this question by applying Kepler's third law, which states that:

"The square of the orbital period of a planet around the Sun is proportional to the cube of the semi-major axis of its orbit"

Mathematically,

$\frac{T^2}{a^3}=const.$

Where

T is the orbital period

a is the semi-major axis of the orbit

In this problem, the question asks what happens if the distance of the Earth from the Sun increases. Increasing this distance means increasing the semi-major axis of the orbit, $a$ : but as we saw from the previous equation, the orbital period of the Earth is proportional to $a$ , therefore as $a$ increases, T increases as well.

Therefore, the period of the orbit would increase.

Learn more about Kepler's third law:

brainly.com/question/11168300

#LearnwithBrainly

5 0

3 years ago