Yes smaller particles are used for energy.
Answer: 100 grams of the parent isotope will remain after one half life.
Explanation:
Mass of the isotope present at initial stage = 
The mass of the parent isotope left after the time ,t=N
Time taken by the samle ,t = 
The half life of the sample :
![\ln[N]=ln[N^o]-\frac{0.693}{t_{\frac{1}{2}}}\times t_{\frac{1}{2}}](https://tex.z-dn.net/?f=%5Cln%5BN%5D%3Dln%5BN%5Eo%5D-%5Cfrac%7B0.693%7D%7Bt_%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%5Ctimes%20t_%7B%5Cfrac%7B1%7D%7B2%7D%7D)
![2=\frac{[N_o]}{[N]}](https://tex.z-dn.net/?f=2%3D%5Cfrac%7B%5BN_o%5D%7D%7B%5BN%5D%7D)
![[N]=\frac{N_o}{2}=\frac{200 g}{2}=100 g](https://tex.z-dn.net/?f=%5BN%5D%3D%5Cfrac%7BN_o%7D%7B2%7D%3D%5Cfrac%7B200%20g%7D%7B2%7D%3D100%20g)
100 grams of the parent isotope will remain after one half life.
Student 4 is correct. The farther the planet is from the Sun, the less pull it exerts on it.
Answer: 0.18
Explanation:
For the alleles, the percentage distribution of each is 'A' (90% = 0.9)
While 'a' (10% = 0.1)
Hence, 0.9 and 0.1 are the respective frequencies of each allele
Now, apply Hardy-Weinberg Equilibrium equation, where heterozygotes are represented by the 2pq term.
Therefore, the number of heterozygous individuals (Aa) is equal to 2pq which equals
2 × 0.9 × 0.1 = 0.18
Thus, the frequency of heterozygote is 0.18, while the percentage distribution in the population is 18%
