Chemistry, 08.01.2020 13:31, Daddysgirl2019

Io, one of jupiter’s moons, is about the size of earth’s moon. io has a radius of about 1.85 x 106 m, and the mean distance between io and jupiter is 4.22 x 108 m. jupiter has a mean radius of 7.15 x 107m, and a mass of 1.90 x 1027kg. if io’s orbit is circular, what is its orbital speed? a. 2.34 x 104 m/s c. 8.7 x 106 m/s b. 1.60 x 104 m/s d. 2.60 x 103 m/s

Answers: 2

Show answers

Answers

This subject is 100% physics.

You need to use Newton's Universal Law of Gravitation and the equation of centripetal accelerationt for the circular motion.

From Newton's Universal Law of Gravitation you can find the acceleration with which Jupiter attracts Io.

This is the formula: F = G*M*m/(r^2)

Where G is the universal constant of gravitation = 6.67 * 10^ -11 N*m^2 / kg^2

M is the mass of JUpiter = 1.90*10^27 kg

r is the distance that separates the center of the two objects, i.e the radius of them both plus the distance between them

r = 4.22*10^8 m + 1.85*10^6 m + 7.15*10^7 m = 4.9535 * 10^8 m

The factor G*M/(r^2) is the gravitational acceleration, which we can calculate now:

6,67 * 10^ - 11 N*m^2 / kg^2 * 1.90*10^27 kg / (4.9535 * 10^8 m)^2 = 0.51648 m/s^2

That is the centripetal acceleration. Call it Ac and use the equation of the circular motion for the centripetal acceleration:

Ac = v^2 / R

Here R = the radius of the orbit, which is the distance that separates the two objects plus the radius of the two objects, so you can solve for v:

v^2 = Ac*R = 0.51648 m/s^2 * (4.9535*10^8m) = 2.558 * 10^8 m^2/s^2

=> v = 1.599 * 10^4 m/s

And the answer is the option b. 1.60 * 10^4 m/s

.

answered: Guest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

answered: Guest

Bthe s the

Other questions on the subject: Chemistry

Chemistry, 21.06.2019 17:30, kev71

You are performing an experiment in a lab to attempt a new method of producing pure elements from compounds. the only problem is that you do not know what element will form. by your previous calculations you know that you will have 6.3 moles of product. when it is complete, you weigh it and determine you have 604.4 grams. what element have you produced?

Answers: 1

continue

Chemistry, 22.06.2019 10:10, babyphoraaaaa

For the reaction, 4 a(g) + 3 b(g) => 2 c(g), the following data were obtained at constant temperature. experiment initial[a],mol/l initial [b],mol/l initial rate, m/min 1 0.200 0.150 5.00 2 0.400 0.150 10.0 3 0.200 0.300 10.0 4 0.400 0.300 20.0 which of the following is the correct rate law for the reaction? 1. rate = k[a]2[b]2 2. rate = k[a][b] 3. rate = k[a]2[b] 4. rate = k[a][b]2

Answers: 3

continue

Chemistry, 22.06.2019 15:50, Edwardwall

Elements in group 2 are all called alkaline earth metals. what is most similar about the alkaline earth metals?

Answers: 1

continue

Chemistry, 22.06.2019 22:30, kiera2599

3.09 lab: reaction of metals 1 which combinations of substances resulted in a chemical change? for each metal that participated in a chemical change, write the type of metal it is, based on your examination of the periodic table. were there any metallic compounds that did not react with either the acid or the base? write the type of metal, based on your examination of the periodic table. make a general statement about the reactivity of the metals in this experiment.

Answers: 1

continue

Do you know the correct answer?

Io, one of jupiter’s moons, is about the size of earth’s moon. io has a radius of about 1.85 x 106 m...