Answer:
counting tree rings can indicate age
Explanation:
hope i helped at least a little bit :)
<h2>
Answer: 1.252</h2>
Explanation:
We are given this equation and we need to find the value of
:
(1)
Firstly, we have to clear
:
(2)
Applying<u> Natural Logarithm</u> on both sides of the equation (2):
(3)
(4)
According to the Natural Logarithm rules
, so (4) can be written as:
(5)
Finally:
Answer:
The orbital velocity 
The period is 
Explanation:
Generally centripetal force acting ring particle is equal to the gravitational force between the ring particle and the planet , this is mathematically represented as

=> 
Here G is the gravitational constant with value 
is the mass of with value 
r is the is distance from the center of the to the outer edge of the A ring
i.e r = R + D
Here R is the radius of the planet with value 
D is the distance from the equator to the outer edge of the A ring with value 
So

=> 
So
=> ![w = \sqrt{ \frac{ 6.67*10^{-11}* 5.683*10^{26}}{[1.4*10^{8}]^3} }](https://tex.z-dn.net/?f=w%20%3D%20%5Csqrt%7B%20%5Cfrac%7B%206.67%2A10%5E%7B-11%7D%2A%20%205.683%2A10%5E%7B26%7D%7D%7B%5B1.4%2A10%5E%7B8%7D%5D%5E3%7D%20%7D)
=> 
Generally the orbital velocity is mathematically represented as

=> 
=> 
Generally the period is mathematically represented as

=> 
=> 
Answer:
The orbital velocity 
The period is 
Explanation:
Generally centripetal force acting ring particle is equal to the gravitational force between the ring particle and the , this is mathematically represented as

Answer:
(D) 4
Explanation:
The percentage error in each of the contributors to the calculation is 1%. The maximum error in the calculation is approximately the sum of the errors of each contributor, multiplied by the number of times it is a factor in the calculation.
density = mass/volume
density = mass/(π(radius^2)(length))
So, mass and length are each a factor once, and radius is a factor twice. Then the total percentage error is approximately 1% +1% +2×1% = 4%.
_____
If you look at the maximum and minimum density, you find they are ...
{0.0611718, 0.0662668} g/(mm²·cm)
The ratio of the maximum value to the mean of these values is about 1.03998. So, the maximum is 3.998% higher than the "nominal" density.
The error is about 4%.
_____
<em>Additional comment</em>
If you work through the details of the math, you will see that the above-described sum of error percentages is <em>just an approximation</em>. If you need a more exact error estimate, it is best to work with the ranges of the numbers involved, and/or their distributions.
Using numbers with uniformly distributed errors will give different results than with normally distributed errors. When such distributions are involved, you need to carefully define what you mean by a maximum error. (By definition, normal distributions extend to infinity in both directions.) While the central limit theorem tends to apply, the actual shape of the error distribution may not be precisely normal.