A 0.5 kg brick is dropped from a height of 6.0 m. It hits the ground and comes to rest. (a) What is the magnitude of the impulse exerted by the ground on the brick? N·s (b) If it takes 0.0013 s from the time the brick first touches the ground until it comes to rest, what is the magnitude of the average force exerted by the ground on the brick?

a). Impulse = 32.373 Ns

b). Force = 2490 N

Explanation:

Impulse = force * time

Force = Mgh

But time is unknown.

From

V² = U² + 2as

But V = at

(at)²= 0² + 2as

But a = 9.81. ( Acceleration of free fall)

t² =( 2*9.81*6)/96.2361

t = 1.1 sec

Force = Mgh

Force = 0.5*6*9.81

Force = 29.43 N

Impulse = 29.43*1.1

Impulse = 32.373 Ns

But if it takes 0.013 sec ...

Force = 32.373/0.013

Force = 2490 N

i think the answer is b

  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