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icang [17]
3 years ago
9

Suppose you have a rock that, when it solidifies, contains 1 microgram of a radioactive isotope. How much of this isotope remain

s after five half-lives? a) none b) 1/8 microgram c) 1/16 microgram d) 1/32 microgram e) Cannot be determined without knowing the daughter isotope.
Physics
1 answer:
algol133 years ago
7 0

Answer:

d) 1/32 microgram

Explanation:

First half life is the time at which the concentration of the reactant reduced to half.

Second half reaction is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/4.

Third half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/8.

Forth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/16.

Fifth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/32.

The initial mass of the sample = 1 microgram

After 5 half-lives, the mass should reduce to 1/32 of the original.

So the concentration left = 1/32 of 1 microgram = 1/32 microgram

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A 19.0-g bullet embeds itself in a 8.0-kg wooden block, which rests on a horizontal frictionless surface and is attached to a ho
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if k is the force constant and x is the maximum compression distance, then:

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and, the kinetic energy system is given by:

K = 1/2×m×(v^2)

if Ki is the initial kinetic energy of the system, Ui is the initial kinetic energy of the system and Kf and Uf are final kinetic and potential energy respectively then, According to energy conservation:

initial energy  = final energy

            Ki +Ui = Kf +Uf

Ui = 0 J and Kf = 0J

                  Ki = Uf

   1/2×m×(v^2) = 1/2×k×(x^2)

         m×(v^2) = k×(x^2)

                v^2 = k×(x^2)/m

                       = (500)×((21×10^-2)^2)/(19×10^-3 + 8)

                       = 2.75

                    v = 1.66 m/s

the v is the final velocity of the bullet block system, if m1 is the mass of bullet and M is the mass of the block and v1 is the initial velocity of the bullet while V is the initial velocity of the block, then by conservation linear momentum:

m1×v1 + M×V = v×(m1 + M) but V = 0 because the block is stationary, initially.

          m1×v1 = v×(m1 + M)

                v1 =  v×(m + M)/(m1)

                    =  (1.66)×(19×10^-3 + 8)/(19×10^-3)

                    = 699.86 m/s

                    ≈ 700 m/s

Therefore, the velocity of the bullet just before it hits the block is 700 m/s.  

5 0
3 years ago
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