Ablock is released from rest and slides down an incline. the coefficient of sliding friction is 0.38 and the angle of inclination is 60.0°. use energy considerations to find how fast the block is sliding after it has traveled a distance of 38.7 cm along the incline.

2.27m/s

Explanation:

The coefficient of sliding friction = 0.38, the incline angle = 60o

Coefficient of sliding friction = frictional force/ force of normal (fn)

The normal force = mgcos60 where m is the mass of the body and g is acceleration due to gravity

Frictional force = coefficient of friction*fn = 0.38*9.81*m*cos 60 = 1.8639m

Force tending to push the body down the incline plane = mgsin60 = 8.496m

Net force on the body = force pushing downward - Frictional force

Net force = 8.496m - 1.8639m = 6.632m

Kinetic energy of the body = 1/2 mv^2 = work done by the net force pushing the body down = net force * distance travelled

1/2mv^2 = 6.632m * 0.387in meters

Cancel mass on both side leaves 1/2v^2 = 6.632*0.387

V^2 = 2.566584*2 = 5.133

V = √5.133 = 2.27m/s

  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