F=mg=Gm1m2/r^2
g=Gm2/r^2
g=2Gm2/(2r)^2=2Gm2/4r^2=Gm2/2r^2
So since there is half times the gravity on this unknown planet that has twice earth's mass and twice it's radius, then the person can jump twice as high. 1.5*2= 3m high
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 new magnitude of the force of attraction will be 6 times the original force of attraction
<h3>How to determine the initial force </h3>
- Mass 1 = m₁
- Mass 2 = m₂
- Gravitational constant = G
- Distance apart = r
- Initial force (F₁) = ?
F = Gm₁m₂ / r²
F₁ = Gm₁m₂ / r²
<h3>How to determine the new force </h3>
- Mass 1 = 2m₁
- Mass 2 = 3m₂
- Gravitational constant = G
- Distance apart (r) = r
- New force (F₂) =?
F = Gm₁m₂ / r²
F₂ = G × 2m₁ × 3m₂ / r²
F₂ = 6Gm₁m₂ / r²
But
F₁ = Gm₁m₂ / r²
Therefore
F₂ = 6Gm₁m₂ / r²
F₂ = 6F₁
Thus, the new magnitude of the force of attraction will be 6 times the original force of attraction
Learn more about gravitational force:
brainly.com/question/21500344
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