Answer:
Cracking of an egg is a physical change since the egg and the stuff inside does not change but the shape or appearance of the shell changes.
Explanation:
Hope it helps
Answer: the image distance is -18, 28 cm this means behind of the concave mirror. The image size is 2.2 higher that the original so it has 8.8 cm with the same orientation as original and it is a virtual imagen.
Explanation: In order to sove the imagen formation for a concave mirror we have to use the following equation:
1/p+1/q=1/f where p and q represents the distance to the mirror for the object and imagen, respectively. f is the focal length for the concave mirror.
replacing the values we obtain:
1/8.3+1/q=1/15.2
so 1/q=(1/15.2)-(1/8.3)=-54.7*10^-3
then q=-18.28 cm
The magnification is given by M=-q/p=-(-18,28)/8.3= 2.2
We also add a picture to see the imagen formation for this case.
Definitely not me!!!! (unless it's magnetism or circuits)
Answer:
a)143.8 decays/minute
b)0.41 decays/minute
Explanation:
From;
0.693/t1/2 = 2.303/t log (Ao/A)
Where;
t1/2=half-life of C-14= 5670 years
t= time taken to decay
Ao= activity of a living sample
A= activity of the sample under study
a)
0.693/5670 = 2.303/1000 log(162.5/A)
1.22×10^-4 = 2.303×10^-3 log(162.5/A)
1.22×10^-4/2.303×10^-3 = log(162.5/A)
0.53 × 10^-1 = log(162.5/A)
5.3 × 10^-2 = log(162.5/A)
162.5/A = Antilog (5.3 × 10^-2 )
A= 162.5/1.13
A= 143.8 decays/minute
b)
0.693/5670 = 2.303/50000 log(162.5/A)
1.22×10^-4 = 4.61×10^-5 log(162.5/A)
1.22×10^-4/4.61×10^-5 = log(162.5/A)
0.26 × 10^1 = log(162.5/A)
2.6= log(162.5/A)
162.5/A = Antilog (2.6 )
A= 162.5/398.1
A= 0.41 decays/minute