Suppose earth is a soid sphere which will attract the body towards its centre.So, acc. to law of gravitation force on the body will be,
F=G*m1m2/R^2
but we now that F=ma
and here accleration(a)=accleration due to gravity(g),so
force applied by earth on will also be mg
replace above F in formula by mg and solve,
F=G*mE*m/R^2 ( here mE is mass of earth and m is mass of body)
mg=G*mEm/R^2
so,
g =G*mE/R^2
Answer:
The "solid force"? ... The direction of the force always seems to be coming out of the solid surface. A direction which is perpendicular to the plane of a surface is said to be normal. The force that a solid surface exerts on anything in the normal direction is called the normal force.
Explanation:
i think i hope this helps
Answer:
Strong nuclear force is 1-2 order of magnitude larger than the electrostatic force
Explanation:
There are mainly two forces acting between protons and neutrons in the nucleus:
- The electrostatic force, which is the force exerted between charged particles (therefore, it is exerted between protons only, since neutrons are not charged). The magnitude of the force is given by

where k is the Coulomb's constant, q1 and q2 are the charges of the two particles, r is the separation between the particles.
The force is attractive for two opposite charges and repulsive for two same charges: therefore, the electrostatic force between two protons is repulsive.
- The strong nuclear force, which is the force exerted between nucleons. At short distance (such as in the nucleus), it is attractive, therefore neutrons and protons attract each other and this contributes in keeping the whole nucleus together.
At the scale involved in the nucleus, the strong nuclear force (attractive) is 1-2 order of magnitude larger than the electrostatic force (repulsive), therefore the nucleus stays together and does not break apart.
The height at time t is given by
h(t) = -4.91t² + 34.3t + 1
When the ball reaches maximum height, its derivative, h'(t) = 0.
That is,
-2(4.91)t+34.3 = 0
-9.82t + 34.3 = 0
t = 3.4929 s
Note that h''(t) = -9.82 (negative) which confirms that h will be maximum.
The maximum height is
hmax = -4.91(3.4929)² + 34.3(3.4929) + 1
= 60.903 m
Answer:
The ball attains maximum height in 3.5 s (nearest tenth).
The ball attains a maximum height of 60.9 m (nearest tenth)