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tekilochka [14]
2 years ago
11

What is the number N0 of 99mTc atoms that must be present to have an activity of 15mCi?

Physics
1 answer:
svetoff [14.1K]2 years ago
4 0
Base on my research, within 2 hours you have a number of atoms which remain. 
N= N0*2^(-t/6.020 = N= N0*2^-0.33223= 07943 N0

So, the number of atoms that are being disintegrated is N0-N=N0*(1-0.79430)=0.2057 N0

It must be equal to 15 mCi = 15*3.7*10^7= 5.55*10^8 atoms 

N0= 5.55*10*8/0.2057  = 2.698*10^9 atoms

Therefore, 2.698*10^9 atoms is the number of N0
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Deffense [45]

Answer:

m=57.65 kg

Explanation:

Given Data

Ricardo mass m₁=80 kg

Canoe mass m₂=30 kg

Canoe Length L= 3 m

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When Canoe was at rest the net total torque is zero.

Let the center of mass is at x distance from the canoe center and it will be towards the Ricardo cause. So the toque around the center of mass is given as

m_{1}(L/2-x)=m_{2}x+m_{2}(L/2-x)

We have to find m₂.To find the value of m₂ first we need figure out the value of.As they changed their positions the center of mass moved to other side by distance 2x.

so

2x=40

x=40/2

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Substitute in the above equation we get

m_{x}=\frac{m_{1}(L/2-x)-m_{2}x }{L/2+x}\\m_{x}=\frac{80(\frac{3}{2}-0.2 )-30*0.2}{3/2+0.2}\\m_{x}=57.65 kg

4 0
3 years ago
Why are many adults and youth in the United states overweight?
Anna11 [10]

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I'd imagine fast food is a lot more popular than most countries over there and easier to get hold of.

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The ratio of the momentum imparted to gun #1 to that imparted to gun #2 is equal to 2 : 1

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