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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The answer would be A because I did the math :)
Answer: 3.75 mg
Explanation:
We need a dose of 2.50 mg per square meter of BSA, and we also know that the BSA of the child is 1.50 m^2, then if the dosage should be 2.50mg for one m^2, for 1,5 m^2 we should administrate an amount of:
D = 1.5 m^2*(2.50mg/m^2) = 1.5*2.5 mg = 3.75 mg
