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ddd [48]
2 years ago
9

In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to

the top of the slope, a 2.50 kg tube, is pulled at a constant speed by a tow rope that maintains a constant tension of 12.0 N along the direction of the slope. How much thermal energy is created in the slope and the tube during the ascent of a 6.31-m-high, 52.3-m-long slope, in Joule
Physics
1 answer:
IrinaVladis [17]2 years ago
4 0

Answer:

≈ 473 J

Explanation:

m ( mass ) = 2.5 kg

T ( tension ) = 12 N

v = constant

h ( height ) = 6.31 m

d ( diameter ) = 52.3 m

<u>Determine how much thermal energy is created </u>

considering that external force acts on the system

ΔE = W    and this can be rewritten as

mgh + ΔEth = W --------- ( 1 )

where ΔEth  = amount of thermal energy created

also ; W = Fd cos ∅,  hence Work done by Tension force (w) = Td cos 0 = Td

back to equation ( 1 )

mgh + ΔEth = Td

ΔEth  = Td - mgh

         = ( 12 * 52.3 ) - ( 2.5 * 9.8 * 6.31 )

         ≈ 473 J

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OverLord2011 [107]

Answer:

a) h = 13,205.4 m

b)  r_f = 2.12 106 m

c)        e% = 0.68%

Explanation:

a) This is an exercise we are asked to use energy conservation,

Starting point. On the surface of Mimas

        Em₀ = K = ½ m v²

Final point. Where the ball stops

       Em_{f} = U = m g h

        Em₀ = Em_{f}

        ½ m v² = m g h

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let's calculate

         h = ½ 41² / 0.0636

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           Em₀ = K + U = ½ m v² - GmM / r_o

final point. Where the ball stops

            Em_{f}= U = - G mM / r_f

          Em₀ = Em_{f}

          ½ m v² - G m M / r_o = - G mM / r_f

In this case all distances are measured from the center of the satellite

         1 / rf = 1 / GM (-½ v² + G M / r_o)

     

let's calculate

         1 / rf = 1 / (6.67 10⁻¹¹ 3.75 10¹⁹) (- ½ 41 2 + 6.67 10⁻¹¹ 3.75 10¹⁹ / 1.98 105)

         1 / r_f = 3,998 10⁻¹¹(-840.5 + 12.63 10³)

          1 / r_f = 4,714 10⁻⁷

          r_f = 1 / 4,715 10⁻⁷

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to measure this distance from the satellite surface

          r_f ’= r_f - r_o

          r_f ’= 2.12 106 - 1.98 105

         r_f ’= 1,922 106 m

c) the percentage difference is

          e% = 13 205.4 / 1,922 106 100

          e% = 0.68%

The estimate of part a is a little low

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Answer:

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galina1969 [7]

Answer:

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The tangential acceleration instead is given by

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Since the girl is near the outer edge and the boy is closer to the centre, the value of r for the girl is larger than for the boy, so the girl has greater tangential acceleration.

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