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
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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.
Answer:
see below
Explanation:
a. 0.1886 x 12
=2.2632
This has 2 sig figures so the answer can only have 2 sig figures
2.3
b. 2.995 - 0.16685
=2.82815
The most accurate in the problem is to thousands place so our answer can only be accurate to the thousands place
2.828
c. 910 x 0.18945=172.3995
The least number of significant figures is 3 so the answer can only have 3 significant figures
172
Answer:
The speed of the large cart after collision is 0.301 m/s.
Explanation:
Given that,
Mass of the cart, 
Initial speed of the cart, 
Mass of the larger cart, 
Initial speed of the larger cart, 
After the collision,
Final speed of the smaller cart, 
(as its recolis)
To find,
The speed of the large cart after collision.
Solution,
Let 
is the speed of the large cart after collision. It can be calculated using conservation of momentum as :
So, the speed of the large cart after collision is 0.301 m/s.
Answer:
Explanation:
To solve this problem, we can use the following suvat equation:
where
is the vertical displacement of the frog
is the initial vertical velocity
t is the time
a is the acceleration
We have chosen this formula because apart from , all the other quantities are known. In fact:
is the vertical displacement
t = 2 s is the total time of flight
is the acceleration due to gravity (negative because it is downward)
Therefore, solving for , we find the initial velocity of the frog:
