Answer:
t = 5.59x10⁴ y
Explanation:
To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:
(1)
<em>where 
: is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>
To find A₀ we can use the following equation:
(2)
<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>
From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:
<em>where 
: is the tree's carbon mass, 
: is the Avogadro's number and 
: is the ¹²C mass. </em>
Similarly, from equation (2) λ is:
<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>
So, the initial activity A₀ is:
Finally, we can calculate the time from equation (1):
I hope it helps you!
Apparent magnitude depends mainly on the brightness of the object as seen from an observer on Earth. This is taken into account without the effects of the atmosphere.
Explanation:
In my view, when the Object A is attracted to a Charged object B. Object B should be Negatively or Positively charged. So Object B should be the Opposite charged according to the Object B
Example =
If Object B is Negatively Charged, the Object A should be Positively Charged
If the Object B is Positively Charged, the Object A should be Negatively Charged
Sometimes it can Mix as a Neutral as well
Hope this Helps
Plate tectonics<span>is a </span>scientific theory<span> that describes the large-scale motion of </span>Earth<span>'s </span>lithosphere<span>. This theoretical model builds on the concept of </span>continental drift<span> which was developed during the first few decades of the 20th century. The </span>geoscientific<span> community accepted plate-tectonic theory after </span>seafloor spreading was validated in the late 1950s and early 1960s.<span>The lithosphere, which is the rigid outermost shell of a planet (the crust and upper mantle), is broken up into </span>tectonic plates<span>. </span>
Answer:
The minimum possible coefficient of static friction between the tires and the ground is 0.64.
Explanation:
if the μ is the coefficient of static friction and R is radius of the curve and v is the speed of the car then, one thing we know is that along the curve, the frictional force, f will be equal to the centripedal force, Fc and this relation is :
Fc = f
m×(v^2)/(R) = μ×m×g
(v^2)/(R) = g×μ
μ = (v^2)/(R×g)
= ((25)^2)/((100)×(9.8))
= 0.64
Therefore, the minimum possible coefficient of static friction between the tires and the ground is 0.64.
