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4 years ago

6

The gravitational force of attraction between two students sitting at their desks in physics class is 2.59 × 10−8 N. If one stud

ent has a mass of 31.9 kg and the other has a mass of 30.0kg, how far apart are the students sitting? The universal gravitational constant is 6.673 × 10−11 N · m2/kg^2.

2 answers:

11Alexandr11 [23.1K]4 years ago

4 0

Answer:1.57x10^(-8)m

Explanation:

Force(f)=2.59 x 10^(-8)N

Mass1(M1)=31.9kg

Mass2(M2)=30kg

Gravitational constant(G)=6.673x10^(-11)

Distance apart(d)=?

F=(GxM1xM2)/d^2

2.59x10^(-8)=(6.673x10^(-11)x31.9x30)/d^2

2.59x10^(-8)=(6.39x10^(-8))/d^2

d^2=(6.39x10^(-8))/(2.59x10^(-8))

d^2=2.47x10^(-16)

d=√(2.47x10^(-16))

d=1.57x10^(-8)m

motikmotik4 years ago

3 0

<h2>The distance between students is 2.46 m</h2>

Explanation:

The force of attraction due to Newton's gravitation law is

F = $\frac{Gm_1m_2}{r^2}$

Here G is the gravitational constant

m₁ is the mass of one student

m₂ is the mass of second student .

and r is the distance between them

Thus r = $\sqrt{\frac{Gm_1m_2}{F} }$

If we substitute the values in the above equation

r = $\sqrt{\frac{6.673x10^-^1^1x31.9x30.0}{2.59x10^-^8} }$

= 2.46 m

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