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EleoNora [17]
3 years ago
8

The roller-coaster car shown in fig. 6-41 (h1 = 30 m, h2 = 12 m, h3 = 20 m), is dragged up to point 1 where it is released from

rest. assuming no friction, calculate the speed at points 2, 3, and 4.
Physics
1 answer:
maxonik [38]3 years ago
7 0
<span>Since there is no friction, conservation of energy gives change in energy is zero Change in energy = 0 Change in KE + Change in PE = 0 1/2 x m x (vf^2 - vi^2) + m x g x (hf-hi) = 0 1/2 x (vf^2 - vi^2) + g x (hf-hi) = 0 (vf^2 - vi^2) = 2 x g x (hi - hf) Since it starts from rest vi = 0 Vf = squareroot of (2 x g x (hi - hf)) For h1, no hf Vf = squareroot of (2 x g x (hi - hf)) Vf = squareroot of (2 x 9.81 x 30) Vf = squareroot of 588.6 Vf = 24.26 For h2 Vf = squareroot of (2 x 9.81 x (30 – 12)) Vf = squareroot of (9.81 x 36) Vf = squareroot of 353.16 Vf = 18.79 For h3 Vf = squareroot of (2 x 9.81 x (30 – 20)) Vf = squareroot of (20 x 9.81) Vf = 18.79</span>
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Answer:

The illumination on the book before the lamp is moved is 9 times the illumination after the lamp is moved.

Explanation:

The distance of the book before the lamp is moved, d_{b} = 30 cm

The distance of the book after the lamp is moved, d_{a} = 90 cm

Illumination can be given by the formula, E = \frac{P}{4 \pi d^{2} }

Illumination before the lamp is moved, E_{b} = \frac{P}{4 \pi d_{b} ^{2} }

Illumination after the lamp is moved, E_{a} = \frac{P}{4 \pi d_{a} ^{2} }

\frac{E_{a}}{E_{b}} } = \frac{\frac{P}{4 \pi d_{a} ^{2} } }{\frac{P}{4 \pi d_{b} ^{2} } }

\frac{E_{a} }{E_{b} } = \frac{d_{b} ^{2} }{d_{a} ^{2}} \\\frac{E_{a} }{E_{b} } = \frac{30^{2} }{90 ^{2}}\\\frac{E_{a} }{E_{b} } =\frac{900 }{8100}\\\frac{E_{a} }{E_{b} } =\frac{1 }{9}

E_{b} = 9E_{a}

The illumination on the book before the lamp is moved is 9 times the illumination after the lamp is moved.

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