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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Answer:
2.36 x 10^6 J
Explanation:
Tc = 0°C = 273 K
TH = 22.5°C = 295.5 K
Qc = heat used to melt the ice
mass of ice, m = 85.7 Kg
Latent heat of fusion, L = 3.34 x 10^5 J/kg
Let Energy supplied is E which is equal to the work done
Qc = m x L = 85.7 x 3.34 x 10^5 = 286.24 x 10^5 J
Use the Carnot's equation


QH = 309.8 x 10^5 J
W = QH - Qc
W = (309.8 - 286.24) x 10^5
W = 23.56 x 10^5 J
W = 2.36 x 10^6 J
Thus, the energy supplied is 2.36 x 10^6 J.
The mineral with Mohs hardness would be scratched because the mineral with Mohs 7 hardness is stronger than the Mohs 5 mineral. Eventually, that mineral would turn into dust if you kept rubbing it.
Answer:
Surface 1 is blacktop, Surface 2 is gravel, and Surface 3 is ice.
Explanation:
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