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Edit Vew History Bookmarks Window Help the fight of a socc.. Soccer Tactics as Science? Learning Jump to Score! 03/4/2018 11:59 PM 086.5/100 Gradebook pts Question 8 of 23 University Physics Froghopper insects have a typical mass of around 12.3 mg and can jump to a height of 558 cm. The takeotftf velocity is achieved as the insect flexes its legs over a distance of approximateily 2.00 mm. Assume that the ump is verical and that the troghopper undergoms constant acceleration while its feet are in contact with the ground. Ignore air resistance What is the acceleration of the insect during the time of the jump (before it leaves the ground)? Number During the jump, how long are the froghopper's feet in contact with the ground? force did the ground exert on the froghopper during the jump? Express your answer as a multiple of the insects weight Number Et Previous Give Up & View Sun 9Check AnswerNeat

Explanation / Answer

mass, m=12.3*10^-3 kg

height, h=55.8 cm

distnace, d=2mm

a)

use,

v=sqrt(2*g*h)

V=sqrt(2*9.8*0.558)

takeooff speed, V=3.307 m/sec

and

to find the acceleration,

v^2 -u^2=2*a*d

3.037^2-0=2*a*2*10^-3

===> a=2.305*10^3 m/sec^2

b)

use,

v=u*a*t

3.307=0+2.305*10^3*t

==> t=1.436*10^-3 sec or 1.436 msec

c)

force on froghopper, F=m*(g+a)

weight, w=m*g

F/w=m*(g+a)/(m*g)

=(g+a)/g

=(9.8+2.305*10^3)/9.8

=236.2

===> F/w=(236.2)

or F=236.2*w

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