Answer:
m =8.81*10^{-6}grams
time t = 52.8 year
Explanation:
GIVEN DATA:
the half life of the CO-60 is, T_1/2 = 5.27 years = 1.663 e+8 s
activity dN/dt = 1 mCi = 3.7 X 10^7 decay/s
activity , 
= ( 3.7 X 10^7 )(1.663*10^8 ) / ln2
= 8.877*10^{16}
Number of moles:
n = N/NA = 8.877*10^{16} / 6.022X10^23 = 1.474*10^{-7} mol
mass of the CO-60 is,
m = n*M = [1.474*10^{-7} mol]*[59.93 grams /mol] = 8.81*10^{-6}grams
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time t = -[T1/2 / ln2]*ln[N/N0]
= - [5.3 years / ln2]*ln[1x10-6/1x10-3]
= 52.8 year
The answer is B because it's 80 percent copper, that's a greater number then 20.
Answer:
Original speed of the mess kit = 4.43 m/s at 50.67° north of east.
Explanation:
Let north represent positive y axis and east represent positive x axis.
Here momentum is conserved.
Let the initial velocity be v.
Initial momentum = 4.4 x v = 4.4v
Velocity of 2.2 kg moving at 2.9 m/s, due north = 2.9 j m/s
Velocity of 2.2 kg moving at 6.8 m/s, 35° north of east = 6.9 ( cos 35i + sin35 j ) = 5.62 i + 3.96 j m/s
Final momentum = 2.2 x 2.9 j + 2.2 x (5.62 i + 3.96 j) = 12.364 i + 15.092 j kgm/s
We have
Initial momentum = Final momentum
4.4v = 12.364 i + 15.092 j
v =2.81 i + 3.43 j
Magnitude
Direction
50.67° north of east.
Original speed of the mess kit = 4.43 m/s at 50.67° north of east.
Average Velocity= displacement/time Av=50/0.50 Av=100
