Ahighway curve of radius 60m is banked at an 18° angle. what is the maximum speed with which a 1500 kg rubber tired car can take this curve without sliding? assume the coefficient of static friction between the tires and the road is 1.0. hint: you must consider carefully which way friction points, up the incline or down the incline. also, remember that for uniform circular motion problems you always want to use radial, tangential, and z axes.

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Explanation:

To solve this problem we need to apply Newton's Second Law to the uniform circular motion, this means, we are going to take in account the radial forces. The forces we have in the x axiss are:

(we do not consider the weight here since it is only exerted in the y axiss) so

,

the friction points toward the center of the curve, this is why the car will not slide,

,

now remembering that

we get

,

(this will be eq 1).

Now, for the y axiss we have:

Where the weight and the vertical component of the friction point downwards.

So

,

,

,

(this will be eq 2).

Now we do eq 1 over eq 2:

,

simplifiying and solving for

,

,

,

,

and finaly

  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

dang that balls went far