A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.540 m/s. The block has a head-on elastic collision with a 40.0 gram block that is initially at rest. Since the collision is head-on, all velocities lie along the same line, both before and after the collision. (a) What is the speed of the 20.0 gram block after the collision

  0.180 m/s

Explanation:

Solving the equations for conservation of momentum and energy for an elastic collision gives ...

  v₁' = ((m₁ -m2)v₁ +2m₂v₂)/(m₁ +m₂) . . . . v₁' is the velocity of m₁ after collision

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Here, we have (m₁, m₂, v₁, v₂) = (20 g, 40 g, 0.540 m/s, 0 m/s).

Substituting these values in to the equation for v₁', we have ...

  v₁' = ((20 -40)(0.540) +2(40)(0))/(20 +40) = (-20/60)(0.540)

  v₁' = -0.180 . . . m/s

The speed of the 20 g block after the collision is 0.180 m/s to the left.

  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

speed

ratio of distance traveled to time interval

net force

vector sum of forces on object

object

source of diverging light rays either luminous or illuminated

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explanation: