Answer:
Easy search it on g o o g l e
Answer:
The answer is "
"
Explanation:
In point i:


If error in the theoretical time period
:



In point ii:

<h3>

</h3>
Answer:

Explanation:
Given the parallex of the star is 0.1 sec.
The distance is inversely related with the parallex of the star. Mathematically,

Here, d is the distance to a star which is measured in parsecs, and P is the parallex which is measured in arc seconds.
Now,

And also know that,

Therefore the distance of the star is
away.
Hmmm. I think the answer is A. If I remember correctly. I had this question before. Sorry if I'm wrong.
Answer:
K = 588.3 N/m
Explanation:
From a forces diagram, and knowing that for the maximum value of K, the crate will try to rebound back up (Friction force will point downward):
Fe - Ff - W*sin(22) = 0 Replacing Fe = K*X and then solving for X:

By conservation of energy:

Replacing our previous value for X and solving the equation for K, we get maximum value to prevent the crate from rebound:
K = 588.3 N/m