Enterprising students set an enormous slip-n-slide (a plastic sheet covered in water to reduce friction) on flat ground. If the slip-n-slide is 250 meters long. a. How small does the average acceleration have to be for a student starting at 5 meters/second to slide to the end?
b. If the acceleration is 0.34 m/s^2, what is the minimum initial speed a student would have to run to make it to the end? Is this speed possible?

Explanation:

a) Using the equation of motion

v² = u²+2as

v is the final velocity = 0m/s

u is the initial velocity = 5m/s

S is the distance = 250m

a is the acceleration

Substitute given values:

0² = 5²+2a(250)

-25 = 500a

a = -25/500

a = -0.05m/s²

b) If a = 0.34m/s²

We need to get the initial speed u

v² = u²+2as

0 = u²+2(0.34)(250)

-u² = 170

u² = -170

u = √-170

The square root of a negative number will give a complex number, hence the speed is not possible

  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

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft