Answer: 86.47 g of carbon-14 must have been present in the sample 11,430 years ago.
Explanation:
Half-life of sample of carbon -14= 5,730 days

Let the sample present 11,430 years(t) ago = 
Sample left till today ,N= 0.060 g

![ln[N]=ln[N]_o-\lambda t](https://tex.z-dn.net/?f=ln%5BN%5D%3Dln%5BN%5D_o-%5Clambda%20t)
![\log[0.060 g]=\log[N_o]-2.303\times 0.00012 day^{-1}\times 11,430 days](https://tex.z-dn.net/?f=%5Clog%5B0.060%20g%5D%3D%5Clog%5BN_o%5D-2.303%5Ctimes%200.00012%20day%5E%7B-1%7D%5Ctimes%2011%2C430%20days)
![\log[N_o]=1.9369](https://tex.z-dn.net/?f=%5Clog%5BN_o%5D%3D1.9369)

86.47 g of carbon-14 must have been present in the sample 11,430 years ago.
Answer:
0.25 gram of neptunium is remaining
Explanation:
First we calculate the no. of half lives passed. For that we have formula:
n = t/T
where,
n = no. of half lives passed = ?
t = total time passed = 8 days (From Monday noon to Tuesday noon of following week)
T = Half Life Period = 2 days
Therefore,
n = 8 days/2 days
n = 4
Now, for the remaining mass of neptunium, we use the formula:
m = (mi)/(2)^n
where,
mi = initial mass of neptunium = 4.00 grams
m = remaining mass of neptunium = ?
Therefore,
m = 4 grams/2⁴
<u>m = 0.25 gram</u>
C because of galvination is sized
Answer:
after 6 second it will stop
he travel 36 m to stop
Explanation:
given data
speed = 12 m/s
distance = 100 m
decelerates rate = 2.00 m/s²
so acceleration a = - 2.00 m/s²
to find out
how long does it take to stop and how far does he travel
solution
we will apply here first equation of motion that is
v = u + at ......1
here u is speed 12 and v is 0 because we stop finally
put here all value in equation 1
0 = 12 + (-2) t
t = 6 s
so after 6 second it will stop
and
for distance we apply equation of motion
v²-u² = 2×a×s ..........2
here v is 0 u is 12 and a is -2 and find distance s
put all value in equation 2
0-12² = 2×(-2)×s
s = 36 m
so he travel 36 m to stop