The friction force between the box and the incline if the box does not slide down the incline will be 0.577
The force preventing sliding against one another of solid surfaces, fluid layers, and material components is known as friction. There are several kinds of friction: Two solid surfaces in touch are opposed to one another's relative lateral motion by dry friction.
Given the box resting on the inclined plane above has a mass of 20kg and the The incline sits at a 30 degree angle
We have to find the friction force between the box and the incline if the box does not slide down the incline
Since the frictional force F₁ must equal or exceed gravitational force F₂ down the incline:
F₁ = F₂
μmgcosΘ = mgsinΘ
μ = (mgsinΘ)/(mgcosΘ)
μ = tanΘ
μ = 0.577
Hence the friction force between the box and the incline if the box does not slide down the incline will be 0.577
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Complete question:
Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength at the midpoint between the two rings ?
Answer:
The electric field strength at the mid-point between the two rings is zero.
Explanation:
Given;
diameter of each ring, d = 10 cm = 0.1 m
distance between the rings, r = 21.0 cm = 0.21 m
charge of each ring, q = 40 nC = 40 x 10⁻⁹ C
let the midpoint between the two rings = x
The electric field strength at the midpoint between the two rings is given as;
Therefore, the electric field strength at the mid-point between the two rings is zero.
C.) In this, number of Hydrogen atoms is 4
