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storchak [24]
2 years ago
11

The illustration in figure below shows a uniform metre rule weighing 30 N pivoted on a wedge placed under the 40 cm mark and car

rying a weight of 70 N hanging 2 from the 10 cm mark. The ruler is balanced horizontally by a weight W hanging from the 100 cm mark. Calculate the value of the weight W.

Physics
1 answer:
Nitella [24]2 years ago
6 0

Answer:

W = 30 N

Explanation:

Applying the summation of torques about the wedge for equilibrium, taking the clockwise direction as negative. Since the ruler is balanced horizontally about the wedge. Therefore, the summation of all torques acting about the wedge must be equal to zero.

(70\ N)(40\ cm - 10\ cm)-(30\ N)(50\ cm-40\ cm)-(W)(100\ cm - 40\ cm) = 0\\W(60\ cm) = (70\ N)(30\ cm)-(30\ N)(10\ cm)\\\\W = \frac{1800\ N.cm}{60\ cm}

<u>W = 30 N</u>

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2 years ago
A car travels a distance of 140 km at 70.0 km/hr. It then travels an additional distance of 60.0 km at 40.0 km/hr. The average s
Natali5045456 [20]

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57.1 km/hr

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2 years ago
A bead slides without friction around a loop the-loop. The bead is released from a height of 17.6 m from the bottom of the loop-
solong [7]

Answer:

 N₁ = 393.96 N   and  N = 197.96 N

Explanation:

In This exercise we must use Newton's second law to find the normal force. Let's use two points the lowest and the highest of the loop

Lowest point, we write Newton's second law n for the y-axis

          N -W = m a

where the acceleration is ccentripeta

          a = v² / r

           

          N = W + m v² / r

          N = mg + mv² / r

         

we can use energy to find the speed at the bottom of the circle

starting point. Highest point where the ball is released

           Em₀ = U = m g h

lowest point. Stop curl down

           Em_{f} = K = ½ m v²

           Emo = Em_{f}

           m g h = ½ m v²

           v² = 2 gh

we substitute

             N = m (g + 2gh / r)

            N = mg (1 + 2h / r)

let's calculate

          N₁ = 5 9.8 (1 + 2 17.6 / 5)

          N₁ = 393.96 N

headed up

we repeat the calculation in the longest part of the loop

          -N -W = - m v₂² / r

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we seek speed with the conservation of energy

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final point. Top of circle with height 2r

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              Em₀ =   Em_{f}

            mgh = ½ m v₂² + 2mgr

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            N = mg (2 (h-r) / r 1) = mg (2h/r  -2 -1)

             N = mg (2h/r  - 3)

            N = 5 9.8 (2 17.6 / 5 -3)

            N = 197.96 N

Directed down

3 0
3 years ago
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