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Lana71 [14]
3 years ago
8

You work at a garden store for the summer, and you lift a 14 kg bag of fertilizer with a force of 227 N.

Physics
1 answer:
nika2105 [10]3 years ago
8 0

Answer:

(a) Acceleration of the bag will be a=16.214m/sec^2  

(B) Weight of the bag will be 137.2 N

Explanation:

We have given mass of the bag m = 14 kg

Force with which bag is lifted = 227 N

(A) According to newtons law we force is equal to F = ma , here m is mass and a is acceleration

So 227=14\times a

a=16.214m/sec^2

(b) Acceleration due to gravity g=9.8m/sec^2

We know that weight is given by W = mg , here m is mass and g is acceleration due to gravity

So weight W=mg=14\times 9.8=137.2N

So weight of the bag will be 137.2 N

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Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

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final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

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Answer:

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Explanation:

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