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baherus [9]
4 years ago
10

You need to repair a broken fence in your yard. The hole in your fence is around 3 meters in length and for whatever reason, the

store you go to has oddly specific width 20cm wood. Each plank of wood costs $16.20, how much will it cost to repair your fence?
Physics
1 answer:
Pachacha [2.7K]4 years ago
6 0

The correct answer is $243

Explanation:

The hole in the fence is 3 meters, this means it is necessary to buy wood that covers this distance. Now, each meter is equal to 100 centimeters, this means 3 meters is equivalent to 300 centimeters ( 100 cm in each meter x 3). Besides this, it is known each plank covers 20cm and costs $16.20. In this context, the next step is to find how many planks are needed. The process is shown below:

300 cm (total width) ÷ 20 cm (width of 1 plank) = 15 planks

This means 15 planks are needed. Finally, fin the total cost

15 planks x $16.20 (cost of 1 plan) = $243

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The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

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Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

F_{pull}=m_{T}.a

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Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u>m_{1}<u>:</u>

The only force acting On the m_{1} box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

m_{1} = \frac{F_{12}}{a}

m_{1} = \frac{1387}{1.516}

m_{1} = 915kg

<u>For </u>m_{2}<u>:</u>

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m_{2} = \frac{F_{23}-F_{12}}{a}

m_{2} = \frac{2304-1387}{1.516}

m_{2} = 605kg

<u>For </u>m_{3}<u>:</u>

m_{3} = m_{T} - (m_{1}+m_{2})

m_{3} = 2384.5-1520.0

m_{3} = 864.5kg

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