Froghopper insects have a typical mass of around 11.1 mg 11.1 mg and can jump to a height of 55.3 cm. 55.3 cm. The takeoff velocity is achieved as the insect flexes its legs over a distance of approximately 2.00 mm. 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 resistance. What is the acceleration of the insect during the time of the jump (before it leaves the ground)?

Acceleration is 26114.500m/s^2.

Explanation:

To answer this question Mathematically let's look at two equations of motion.

(1) Vf=vi=at

(2) y(t)=y0+v0t=(1/2)at^2

we can solve it in two steps

first we will basically solve for vi which requires us to solve for t as well.

second we will solve for acceleration which will also require us to solve for t as well.

in both cases we will use these two equtions of motion.

step (1) solving for vf and t.

frog jums 55.3cm , wth initial velocity = 0 that is two piecies of information in eq(1).  and (2),

that gives us t = 3.3seconds and vf = 32.32ms.

step (2).

solving for accelerations . similarly we now that from attains vf in some time while jumping 2.00mm. that two piecies of informations is enought and substituting in (1) and (2) gives.

a =  26114.500m/s^2  and t = 0.0012376seconds.

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz