Question

the force attraction between two bodies is 20N .calculate the force between them when mass of each body as well as the distance between them are halved

Answer 1

Answer:

Explanation:F ∝ m / r^2

F1 / F2 = (m1 / m2) * (r2 / r1)^2

F2 = F1 * (r1 / r2)^2 * (m2 / m1)

F2 = F1 * (r1 / (r1/2))^2 * (m1/2 / m1)

F2 = F1 * 4/2

F2 = 2 * F1

F2 = 2 * 20

F2 = 40 N

Force of attraction will become 40 N

Answer 2

Answer:

40N

Explanation:

Given:

Force= 20N

To find:

Force

Solution:

The force of attraction is a force that, as a result of an attraction, pulls the body toward it. There are a number of attraction factors in nature. These include electromagnetic force, gravitational force, electrostatic force, and electric force. The well-known force of gravity pulls a body toward an object regardless of distance. We learn a lot about this force and how it works from Newton's theory of gravitation. According to this, any mass present in the cosmos will be drawn to the mass already present.

The most well-known force is certainly gravity. When we think about gravity, we think of an apple falling on Isaac Newton's head. We only notice objects falling to the earth because of gravity. Similar events take place everywhere in the cosmos.

F ∝ $\frac{m}{r^{2} }$

$\frac{F_1}{F_2} =(\frac{m_1}{m_2})(\frac{r_2}{r_1} )^{2} \\F_2=(F_1)(\frac{r_2}{r_1} )^{2}(\frac{\frac{m_1}{2} }{m_1} )\\F_2=F_1(\frac{4}{2} )\\F_2=2(F_1)\\F_2=(2)(20)\\F_2=40N$

calculate the force of attraction between an electron and a body with charge +1.0cand it placed at a distance of 1.5m a part k=9.0*10^19 Nm^2/c^2​

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The force of attraction between two objects of masses 2m amd m is F₁.When the object with mass 2m is replaced with another object of mass m/2 ,the new force of attraction is found to be F₂.The ratio of F₁:F₂ IS:

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