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m = mass of the box
N = normal force on the box
f = kinetic frictional force on the box
a = acceleration of the box
μ = coefficient of kinetic friction
perpendicular to incline , force equation is given as
N = mg Cos30 eq-1
kinetic frictional force is given as
f = μ N
using eq-1
f = μ mg Cos30
parallel to incline , force equation is given as
mg Sin30 - f = ma
mg Sin30 - μ mg Cos30 = ma
"m" cancel out
a = g Sin30 - μ g Cos30
inserting the values
1.20 = (9.8) Sin30 - (9.8) Cos30 μ
μ = 0.44
radiation
when the suns radiation fall on the earth and its objects they receive heat energy and hence get heated. Thus the suns heat reaches the earth by. the process of radiation
Given :
A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.
To Find :
The coefficient of static friction between the box and the plane.
Solution :
Vertical component of force :

Horizontal component of force(Normal reaction) :

Since, box is on the verge of slipping :

Therefore, the coefficient of static friction between the box and the plane is 1.07.
Hence, this is the required solution.
Answer:
Explanation:
(a) The force of gravity is called an attractive force because it is the force (although weak) in which a planetary body or matter uses to attract an object towards itself.
(b) Yes, it does and the formula for force of gravity between any two object is
F = G
where m1 and m2 are masses of the first and second object respectively
r is the distance between the center of the two masses
G is the gravitational constant