The "Standard Atmosphere" is 760 mm Hg .
2.13 times that pressure is (2.13 x 760) = 1,618.8 mm Hg.
<span>To answer this problem, we use balancing of forces: x and y components to determine the tension of the rope.
First, the vertical component of tension (Tsin theta) is equal to the weight of the object.
T * sin θ = mg =</span> 1.55 * 9.81 <span>
T * sin θ = 15.2055
Second, the horizontal component of tension (t cos theta) is equal to the force of the wind.
T * cos θ = 13.3
Tan θ = sin </span>θ / cos θ = 15.2055/13.3 = 1.143
we can find θ that is equal to 48.82.
T then is equal to 20.20 N
In the first case, the force acting on the spring is the weight of the mass:

This force causes a stretching of

on the spring, so we can use these data to find the spring constant:

In the second case, the first mass is replaced with a second mass, whose weight is

And since we know the spring constant, we can calculate the new elongation of the spring: