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pashok25 [27]
2 years ago
12

A factory robot drops a 10 kg computer onto a conveyor belt running at 3.1 m/s. The materials are such that μs = 0.50 and μk = 0

.30 between the belt and the computer. How far is the computer dragged before it is riding smoothly on the belt?
Physics
1 answer:
frutty [35]2 years ago
8 0

Answer:

x = 1.63 m

Explanation:

mass (m) = 10 kg

μk = 0.3

velocity (v) = 3.1 m/s

Assuming that most of the computers weight is applied on the belt instantaneously, we can apply the constant acceleration equation below

x = v^{2}/2a

where a = μk.g , therefore

x = v^{2}/2μk.g

x = (3.1 x 3.1)/(2 x 0.3 x 9.8)

x = 1.63 m

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Drupady [299]

Answer:

V_{a} - V_{b} = 89.3

Explanation:

The electric potential is defined by

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         V_{b} - V_{a} = - ∫ E ds

We substitute

         V_{b} - V_{a} = - ∫ (α + β/ y²) dy

We integrate

          V_{b} - V_{a} = - α y + β / y

We evaluate between the lower limit A  2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m

           V_{b} - V_{a} = - α (0.03 - 0.02) + β (1 / 0.03 - 1 / 0.02)

            V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33

            V_{b} - V_{a} = - 89.3 V

As they ask us the reverse case

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             V_{a} - V_{b} = 89.3

3 0
2 years ago
Calculate the force of Earth’s gravity on a spacecraft 2.00 Earth radii above the Earth’s surface (That would be 3.00 Earth radi
igomit [66]

Answer:

2014.44 N

Explanation:

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\frac{g'}{g}=\left (\frac{R}{r}  \right )^{2}

\frac{g'}{g}=\left (\frac{R}{3R}  \right )^{2}

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F = 2014.44 N

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Answer:

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irinina [24]
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