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:
If a vertical line extending down from an object's CG extends outside its area of support, the object will topple
Explanation:
We can understand better this situation using a diagram with the forces acting on it.
In the attached image we can see that when the gravity center is bouncing outside from the area of the pedestal, the object will be out of balance and will fall.
Well, that's a nice, concise description, but it applies to a
generator, not a motor. A motor does exactly the opposite.
It uses an electric current to produce motion in a magnetic field.
Sadly, the statement is false.
I know i did part a correctly. heres what i did: momentum is conserved: m1 * u - m2 * u = m2 * v or (m1 - m2) * u = m2 * v Also, for an elastic head-on collision, we know that the relative velocity of approach = relative velocity of separation (from conservation of energy), or, for this problem, 2u = v Then (m1 - m2) * u = m2 * 2u m1 - m2 = 2 * m2 m1 = 3 * m2 m1 is the sphere that remained at rest (hence its absence from the RHS), so m2 = 0.3kg / 3 m2 = 0.1 kg b) this part confuses me, heres what i did (m1 - m2) * u = m2 * v (.3kg - .1kg)(2.0m/s) = .1kg * v .4 kg = .1 v v = 4 m/s What my teacher did: (.3g - .1g) * 2.0m/s = (.3g + .1g) * v I understand the left hand side but i dont get the right hand side. Why is m1 added to m2 when m1 is at rest which makes its v = zero?? v = +1.00m/s since the answer is positive, what does that mean? Also, if v was -1.00m/s what would that mean? thanks!
<span>Reference https://www.physicsforums.com/threads/elastic-collision-with-conservation-of-momentum-problem.651261...</span>
Answer:
false, true, true, true, false
Explanation:
