A jack in the box is built around with a spring constant of 70N/m. How much hard does the spring push when you compress it from 40 cm long down to 15 cm long?

17.5N

Explanation:

Given that,

spring constant (k) = 70N/m

Original length (lo) = 40 cm

New length (ln) = 15 cm

extension of spring (e) = new length - original length

i.e e = ln - lo

e = 15 cm - 40 cm

e = -25cm (the negative sign is negligible, just the same magnitude matters)

So, convert 25cm to metres

If 100cm = 1m

25cm = 25/100 = 0.25m

(Note that how hard the spring push refers to the Force, hence, find force.)

Now, apply the Hooke law (F = ke)

Force = spring constant x extension

F = 70N/m x 0.25m

F = 17.5N

Thus, the spring push with a force of 17.5N when compressed.

  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

probably not if net force is zero