Answer:
8. 2.75·10^-4 s^-1
9. No, too much of the carbon-14 would have decayed for radiation to be detected.
Explanation:
8. The half-life of 42 minutes is 2520 seconds, so you have ...
1/2 = e^(-λt) = e^(-(2520 s)λ)
ln(1/2) = -(2520 s)λ
-ln(1/2)/(2520 s) = λ ≈ 2.75×10^-4 s^-1
___
9. Reference material on carbon-14 dating suggests the method is not useful for time periods greater than about 50,000 years. The half-life of C-14 is about 5730 years, so at 65 million years, about ...
6.5·10^7/5.73·10^3 ≈ 11344
half-lives will have passed. Whatever carbon 14 may have existed at the time will have decayed completely to nothing after that many half-lives.
The magnitude of the downward acceleration of the hollow cylinder is 6m/s^2.
Z = I α
T.R =1/2 M (
+
)α
T.R = 1/2M 5
/4 α
T = 5Ma/8
Mg - T = Ma
Mg - 5Ma/8 = Ma
Mg= 5Ma/8 + Ma = 13Ma / 8
acceleration = 8g/13 = 6 m/s^2
The rate at which an object's velocity with respect to time changes is called its acceleration. The direction of the net force imposed on an item determines its acceleration in relation to that force. According to Newton's Second Law, the magnitude of an object's acceleration is the result of two factors working together
The size of the net balance of all external forces acting on that item is directly proportional to the magnitude of this net resultant force; the magnitude of that object's mass, depending on the materials from which it is built, is inversely related to its mass.
Learn more about acceleration here:
brainly.com/question/2303856
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Answer:
The (s) indicates that the state of matter for NaHCO3 is solid.
Explanation:
When a chemical reaction is written, the state of matter for each components of the reactants and products are mentioned in brackets along with their names or formulas.
For example, NaHCO3 has (s) mentioned in the brackets. The s shows that the state of matter for NaHCO3. (l) represents liquid format. (g) represents that the state of matter is gas.
Answer:
m = 14*26 = 364
Explanation:
overall magnification is given as m

mo magnification of objective lens
me magnification of EYE lens
where mo is given as

and me as

d is distant of distinct vision = 25.0 cm for normal eye
fe = focal length of eye piece
focal length of objective lense is 0.140 cm
we know that








m = 14*26 = 364