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.
The answer to your question is Metal
Answer:
<h2>104 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 26 × 4
We have the final answer as
<h3>104 N</h3>
Hope this helps you
Answer:
48.6°
Explanation:
The forward force, F equals the component of the weight along the slope.
So mgsinθ = ma where a = acceleration and θ = angle between the slope and the horizontal.
So a = gsinθ
Since we are given that a = 75%g = 0.75g,
0.75g = gsinθ
sinθ = 0.75
θ = sin⁻¹(0.75)
= 48.6°
Answer:
The time taken is 
Explanation:
From the question we are told that
The length of steel the wire is 
The length of the copper wire is 
The diameter of the wire is 
The tension is 
The time taken by the transverse wave to travel the length of the two wire is mathematically represented as
Where 
is the time taken to transverse the steel wire which is mathematically represented as
![t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ]](https://tex.z-dn.net/?f=t_s%20%20%3D%20l_1%20%2A%20%20%5B%20%5Csqrt%7B%20%5Cfrac%7B%5Crho%20%2A%20%5Cpi%20%2A%20%20d%5E2%20%7D%7B%204%20%2A%20%20T%7D%20%7D%20%5D)
here 
is the density of steel with a value 
So
![t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]](https://tex.z-dn.net/?f=t_s%20%20%3D%2031%20%2A%20%20%5B%20%5Csqrt%7B%20%5Cfrac%7B8920%20%2A%203.142%2A%20%20%281%2A10%5E%7B-3%7D%29%5E2%20%7D%7B%204%20%2A%20%20122%7D%20%7D%20%5D)
And
is the time taken to transverse the copper wire which is mathematically represented as
![t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ]](https://tex.z-dn.net/?f=t_c%20%20%3D%20l_2%20%2A%20%20%5B%20%5Csqrt%7B%20%5Cfrac%7B%5Crho_c%20%2A%20%5Cpi%20%2A%20%20d%5E2%20%7D%7B%204%20%2A%20%20T%7D%20%7D%20%5D)
here 
is the density of steel with a value 
So
![t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]](https://tex.z-dn.net/?f=t_c%20%20%3D%2017%20%2A%20%20%5B%20%5Csqrt%7B%20%5Cfrac%7B7860%20%2A%203.142%2A%20%20%281%2A10%5E%7B-3%7D%29%5E2%20%7D%7B%204%20%2A%20%20122%7D%20%7D%20%5D)
So
