<h3><u>Answer;</u></h3>
Carbon-14 levels in a sample are undetectable after approximately 9 half lives
<h3><u>Explanation;</u></h3>
- <em><u>The half life of Carbon-14 is 5,730 years . Half life is the time taken by a radioactive material to decay by half of its original mass. Therefore, it would take a time of 5730 years for a sample of 100 g of carbon-14 to decay to 50 grams</u></em>
- <em><u>A period of 50,000 years, is equivalent to; </u></em>
<em><u> 50,000÷5,730 </u></em>
<em><u>= 8.73 half lives</u></em>
<em>Which is approximately equal to 9 half lives.</em>
- Therefore, if the age of an object older than 50,000 years cannot be determined by radiocarbon dating, then <em><u>Carbon-14 levels in a sample are undetectable after approximately 9 half lives</u></em>.
Answer:
r = 41.1 10⁹ m
Explanation:
For this exercise we use the equilibrium condition, that is, we look for the point where the forces are equal
∑ F = 0
F (Earth- probe) - F (Mars- probe) = 0
F (Earth- probe) = F (Mars- probe)
Let's use the equation of universal grace, let's measure the distance from the earth, to have a reference system
the distance from Earth to the probe is R (Earth-probe) = r
the distance from Mars to the probe is R (Mars -probe) = D - r
where D is the distance between Earth and Mars
M_earth (D-r)² = M_Mars r²
(D-r) =
r
r (
) = D
r =
We look for the values in tables
D = 54.6 10⁹ m (minimum)
M_earth = 5.98 10²⁴ kg
M_Marte = 6.42 10²³ kg = 0.642 10²⁴ kg
let's calculate
r = 54.6 10⁹ / (1 + √(0.642/5.98) )
r = 41.1 10⁹ m
When the forces acting on a body are balanced, their effect
\on the body's motion is the same as if no forces at all are
acting on it, and its velocity can't change. It continues moving
in a straight line at constant speed (which may be zero).
I am sorry I cant find the answer. I was hoping to find the answer
Answer:

Explanation:
Here two charges are placed at distance "d" apart
now the net value of electric field at some position between two charges will be ZERO
so we will have
electric field due to charge 1 = electric field due to charge 2

Let the position where net field is zero will lie at distance "r" from q1

now we will have

now square root both sides

now we have

so we have
