Answer:
94.1 m
Explanation:
From Coulombs law,
F = Gm1m2/r²................... Equation 1
where F = force, m1 = first mass, m2 = second mass, G = universal constant, r = distance of separation.
Make r the subject of the equation,
r = √(Gm1m2/F)................. Equation 2
Given: F = 7×10² N, m1 = 15×10⁷ kg, m2 = 62×10⁷ kg,
Constant: G = 6.67×10⁻¹¹ Nm²/kg²
Substitute into equation 2
r = √( 6.67×10⁻¹¹×15×10⁷×62×10⁷/7×10²)
r √(886.16×10)
r √(88.616×10²)
r = 9.41×10
r = 94.1 m.
Hence the distance of separation = 94.1 m
I’m not too sure but I think it’s 8,91 m/s2
Answer:
Magnitude of static friction force is 70 sin40° = 44.99 N.
No, it is not necessary that it is maximum static friction.
Normal force is equal to 70 cos40° = 53.62 N.
Explanation:
We apply newton law of moton equation along the plane and perpendicular to plane;
Along the plane,
70 sin 40° = 
---------------(1)
70 cos 40° = N --------------(2)
= μN -----------------(3)
So, it depends on the value of μ that the friction is maximum or not .
