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BigorU [14]
2 years ago
14

What is the final position of the object if its initial position is x = 0.40 m and the work done on it is equal to 0.21 J? What

is the final position of the object if its initial position is x = 0.40 m and the work done on it is equal to -0.19 J?

Physics
1 answer:
r-ruslan [8.4K]2 years ago
4 0

Answer:

a) Final position is x = 0.90 m

b) Final position is x = 0.133 m

Explanation:

The workdone between two points is usually approximated as the area under the force-distance curve between those two points.

From the graph,

As at the initial position, x = 0.40 m and the corresponding F = 0.8 N,

The area from that point onwards up to the end of that particular bar = 0.8 (0.5 - 0.4) = 0.08 J

The next bar has force = 0.4 N and the width of the bar = (0.75 - 0.50) = 0.25 m

Work done under this bar = 0.4 × 0.25 = 0.1 J

Total work done from the starting position up to this point now = 0.08 + 0.1 = 0.18 J, still less than 0.21 J

So, the final position has to be on the last bar. Let the position be x. The force on the last bar = 0.2 N

0.21 = 0.18 + 0.2 (x - 0.75)

0.03 = 0.2x - 0.15

0.2x = 0.18

x = 0.9 m

Therefore, the final position of the object, to do 0.21 J worth of work, starting from x = 0.4 m is 0.90 m.

b) For this part, negative work is done, this means, we will move in the negative direction to try and trace this total work done.

From the starting point where the initial position is 0.40 m, the force here is 0.80 N

The workdone under this bar to the left is

The workdone = 0.8 (0.25 - 0.4) = - 0.12 J

Since we're tracing -0.19 J, the final position has to be on the last bar (on the left), Let the position be x. The force on the last bar on the left (could also be referred to as the first bar) = 0.60 N

- 0.19 = -0.12 + 0.6 (x - 0.25)

-0.07 = 0.6x - 0.15

0.6x = 0.08

x = (0.08/0.6) = 0.133 m

Therefore, the final position of the object, after doing -0.19 J worth of work, starting from x = 0.4 m is 0.133 m.

Hope this Helps!!!

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2 years ago
In ideal flow, a liquid of density 850 kg/m3 moves from a horizontal tube of radius 1.00 cm into a second horizontal tube of rad
Crank

Answer:

a)   Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1] , b) Q = 3.4 10⁻² m³ / s , c)      Q = 4.8 10⁻² m³ / s

Explanation:

We can solve this fluid problem with Bernoulli's equation.

         P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

With the two tubes they are at the same height y₁ = y₂

        P₁-P₂ = ½ ρ (v₂² - v₁²)

The flow rate is given by

         A₁ v₁ = A₂ v₂

         v₂ = v₁ A₁ / A₂

We replace

         ΔP = ½ ρ [(v₁ A₁ / A₂)² - v₁²]

         ΔP = ½ ρ v₁² [(A₁ / A₂)² -1]

Let's clear the speed

         v₁ = √ 2ΔP /ρ[(A₁ / A₂)² -1]

The expression for the flow is

           Q = A v

           Q = A₁ v₁

           Q = A₁ √ 2ΔP / rho [(A₁ / A₂)² -1]

The areas are

            A₁ = π r₁

            A₂ = π r₂

We replace

        Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1]

Let's calculate for the different pressures

      r₁ = d₁ / 2 = 1.00 / 2

      r₁ = 0.500 10⁻² m

      r₂ = 0.250 10⁻² m

b) ΔP = 6.00 kPa = 6 10³ Pa

      Q = π 0.5 10⁻² √(2 6.00 10³ / (850 (0.5² / 0.25² -1))

       Q = 1.57 10⁻² √(12 10³/2550)

        Q = 3.4 10⁻² m³ / s

c) ΔP = 12 10³ Pa

        Q = 1.57 10⁻² √(2 12 10³ / (850 3)

         Q = 4.8 10⁻² m³ / s

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Answer:

0.50m/s

Explanation:

Average velocity is the change in displacement of a body with respect to time.

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∆S = 100m - 70m

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<h3>4.53 kgm/s</h3>

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