Which meteorites are the most useful for defining the age of the solar system? Why?

Physics

1 answer:

scoundrel [369]3 years ago

3 0

Answer:

meteorite is a piece of interplanetary debris that lives its fiery drops during a through the earth's atmosphere and strikes the surface of the earth.

Explanation:

the meteorites which are most useful for the determination of the age of the solar system are the primitive meteorites. they consist light of colored or grey silicates mixed with metallic grains. the parent bodies of these meteorites are also mostly believed to be pieces asteriods left after they formed in the solar system.

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Addition of _____________ to pure water causes the least increase in conductivity. A) weak bases B) acetic acid C) ionic compoun

geniusboy [140]

Answer:

Explanation:

Water in itself is a bad conductor of electricity. Any compound that dissociates in water into ions and charged molecules, will increase conductivity of water. In this case ionic compounds weak basis and acids will put in free ions into the water. The ions can pass electricity because they are attracted to respective poles of electricity depending on their charges. In water, these ions are free moving unlike when they are immobilized in their lattice in solid form.

Organic compounds are mainly made of covalent bonds (around carbon) hence do not dissociate in water.

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3 years ago

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Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur

BARSIC [14]

Answer:

$v_A=7667m/s\\\\v_B=7487m/s$

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

$F_g=\frac{GMm}{R^{2} }$

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

$F_c=m\frac{v^{2}}{R}$

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

$\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}$

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So $R_A=6774km=6.774*10^6m$ and $R_B=7103km=7.103*10^6m$ (Since $R_{earth}=6371km$ ). Then, we get:

$v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s$