In an experiment designed to determine the velocity of a bullet fired by a certain gun, a wooden block of mass M = 530.0 g is supported only by its edges, and a bullet of mass m = 7.25 g is fired vertically upward into the block from close range below. After the bullet embeds itself in the block, the block and the bullet are measured to rise to a maximum height of 30.0 cm above the block's original position. What is the speed of the bullet just before impact? m/s

Velocity of bullet = 179.33 m/s

Explanation:

Given:

Mass of block(M) = 530g

Mass of bullet(m) = 7.25g

Height (h) = 30 cm = 0.30 m

Find:

Speed of bullet before impact.

Computation:

After impact velocity (v) =                [∵ g = 9.8]

After impact velocity (v) = 2.42 m/s

Law of conservation .

(M+m)v = M(velocity of block) + m(Velocity of bullet)

(530+7.25)(2.42) = 530(0) + 7.25(Velocity of bullet)

Velocity of bullet = 179.33 m/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

think its the third option