The largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.
The given parameters;
- <em>distance between the two black holes, r = 10 AU = 1.5 x 10¹² m</em>
- <em>gravitational force between the two black holes, F = 6.9 x 10²⁵ N.</em>
- <em>combined mass of the two black holes = 5.20 x 10³⁰ kg</em>
The product of the two masses is calculated from Newton's law of universal gravitational as follows;
The sum of the two masses is given as;
m₁ + m₂ = 5.2 x 10³⁰ kg
m₂ = 5.2 x 10³⁰ kg - m₁
The first mass is calculated as follows;
m₁(5.2 x 10³⁰ - m₁) = 2.328 x 10⁶⁰
5.2 x 10³⁰m₁ - m₁² = 2.328 x 10⁶⁰
m₁² - 5.2 x 10³⁰m₁ + 2.328 x 10⁶⁰ = 0
<em>solve the quadratic equation using formula method</em>;
a = 1, b =- 5.2 x 10³⁰, c = 2.328 x 10⁶⁰
The second mass is calculated as follows;
m₂ = 5.2 x 10³⁰ kg - m₁
m₂ = 5.2 x 10³⁰ kg - 4.7 x 10³⁰ kg
m₂ = 5 x 10²⁹ kg
or
m₂ = 5.2 x 10³⁰ kg - 4.9 x 10²⁹ kg
m₂ = 4.7 x 10³⁰ kg
Thus, the largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.
Learn more here:brainly.com/question/9373839
The second speech bubble to the presidential branch, the third to the judicial branch, and the first to the legislative branch.
Answer:
D: When one bulb burns out, all the other lights stay on
Explanation:
In series combination of light bulbs same current must flow through all the bulbs and hence if one bulb is burn out then current through all bulbs tripped to zero and all bulbs will turn off.
Now in parallel combination all bulbs are connected parallel to the source of energy due to which the bulbs will remain in circuit if any one bulb is burn out.
So here if we used combined circuit of the bulbs i.e. parallel then in that case if one of the bulb is burn out then it must show that rest of the other bulbs must glow.
so here correct answer would be
D: When one bulb burns out, all the other lights stay on
Friction the force that oppose motion so it unhelpful where all the motion of an object is desired.that is what i know.