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History Bookmarks Develop Window Help saplinglearning.com Apple Cloud Yahoo Bing Google Wikipedia Googlo Yahoo Student-Solu..81118217368 it is ethical...ature Medicine pling Learning Macmillan Learning University of California, Merced-PHYS 18-Spring18 Jump to... Gradebook 3/4/2018 11:59 PM 82/ 100 () 3/4/2018 08:27 PM Print Calculator Penoic Table Queston 8 of 23 Mapm University Physics preserted tby Saping Leaning Froghopper insects have a typical mass of around 12.1 mg and can jump to a height of 55.8 cm. The takeoff velocity is achieved as the insect flexes its legs over a distance of approximatcly 2.00 mm. Assume that the jump is vertical and that the froghopper undergoes constant acceleration while its feet are in contact with the ground. Ignore air resistanoe. What is the acceleration of the insect during the time of the jump (before it leaves the ground)? Number In During the jump, how long are the froghopper's foet in contact with the ground? ms What force did the ground exert on the froghopper during the jump? Express your answer as a multiple of the insect's weight O Preveus Give up & View Solution eCheck Answer ONed aEt

Explanation / Answer

Impulse vector = Î = Ft = mv = mv

the insects's velocity in time t changes from 0 to v

insect's weight = mg

since the max height achieved = 55.8 cm = 0.558 m = v²/(2g)

=> v² = 19.62 x 0.558 => v = 3.31 m/s (upwards)

=> mv = m[3.31 - 0] = 3.31m = Ft (where m is the mass) --------> (1)

also s = v(t)

=> 2 x 10 ³ m = ½[0 + 3.31]t

=> t = 2 x 10 ³/1.65 = .00121 s = 1.21 x 10 ³ s = 1.21 ms (answer to 2nd part)

=> Ft = F[1.21 x 10 ³] = 3.31[12.1 x 10 ]   (from eqn (1) )

=> F = 3.31[12.1 x 10 ] / [1.21 x 10 ³] = 0.03313 N

F can be equated to 3.31m/ t,, (see eqn 1)

F = 3.31m / t = 2735.5m

Multiply and divide by g = 9.81

F = 2735.5xmg/9.81 = 278.85mg

mg is the weight

Thus

F= 278.85 times the weight (answer to 3rd part)

a = F/m = 2735.5m/m = 2.735 x 10³ m/s² (Answer to part 1)

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