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:
a) k = 891.82 N/m
b) e = 0.0143 m = 1.43 cm
c) W = 5.02 J
Explanation:
Step 1: Data given
Mass = 2.60 kg
the spring stretches 2.86 cm = 0.0286
Step 2: What is the force constant of the spring?
Force constant, k = force applied / extension produced
k = (2.60kg * 9.81N/kg) / 0.0326 m
k = 891.82 N/m
b) If the 2.60-kg object is removed, how far will the spring stretch if a 1.30-kg block is hung on it
Extension = F/k = (1.30 kg * 9.81) / 891.82 = 0.0143 m = 1.43 cm
Half the mass means half the extension
c) How much work must an external agent do to stretch the same spring 7.50 cm from its unstretched position?
W = average force used * distance
W = 1/2 * k*e * e = 1/2 k*e²
W = 1/2 * 891.82 * (0.075)² = W = 5.02 J
A scalar quantity is one in which there is no direction given.
30 Lbs would be the answer, since there is no indicator of direction.
Answer:
178 cm3
Explanation:
We know that density is expresses as mass per unit volume hence
Density= mass/volume
Making volume the subject then
Volume=mass/density
Substituting mass with 0.00296 g and density with 0.00001663 g/cm3 then
Volume=0.00296/0.00001663=177.99158147925 cm3
Rounding off, the volume is approximately 178 cm3