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marusya05 [52]
3 years ago
9

5. A car stops in 20 m. If it has an acceleration of-6m/s?,

Physics
1 answer:
shtirl [24]3 years ago
5 0

Answer:

15.49 m/s

Explanation:

We know that the stopping distance is ,

=> s = u²/2(-a)

So that ,

=> 20m = u² / 2* 6m/s²

=> u² = 240m²/s²

=> u = √240 m²/s²

=> u = 15.49 m/s

<em>*</em><em>*</em><em>Edits</em><em> </em><em>are </em><em>welcomed</em><em>*</em><em>*</em>

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d) v1 = v2 = v3

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A simple hydraulic lift is made by fitting a piston attached to a handle into a 3.0-cm diameter cylinder. The cylinder is connec
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Answer:

Approximately 3.1 \times 10^4 \; \rm N (assuming that the acceleration due to gravity is g = 9.81\; \rm kg \cdot N^{-1}.)

Explanation:

Let A_1 denote the first piston's contact area with the fluid. Let A_2 denote the second piston's contact area with the fluid.

Similarly, let F_1 and F_2 denote the size of the force on the two pistons. Since the person is placing all her weight on the first piston:

F_1 = W = m \cdot g = 50\; \rm kg \times 9.81 \; \rm kg \cdot N^{-1} =495\; \rm N.  

Since both pistons fit into cylinders, the two contact surfaces must be circles. Keep in mind that the area of a square is equal to \pi times its radius, squared:

  • \displaystyle A_1 = \pi \times \left(\frac{1}{2} \times 3.0\right)^2 = 2.25\, \pi\;\rm cm^{2}.
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By Pascal's Law, the pressure on the two pistons should be the same. Pressure is the size of normal force per unit area:

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For the pressures on the two pistons to match:

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F_1, A_1, and A_2 have all been found. The question is asking for F_2. Rearrange this equation to obtain:

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6 0
3 years ago
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