It is the branch of science, in which we study different phenomena of atmosphere including climate and weather.
Answer:
The spring constant = 104.82 N/m
The angular velocity of the bar when θ = 32° is 1.70 rad/s
Explanation:
From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:


Also;

Thus;

where;
= deflection in the spring
k = spring constant
b = remaining length in the rod
m = mass of the slender bar
g = acceleration due to gravity


Thus; the spring constant = 104.82 N/m
b
The angular velocity can be calculated by also using the conservation of energy;






Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s
The answer would be solubility
Answer:
966 mph
Explanation:
Using as convention:
- East --> positive x-direction
- North --> Positive y-direction
The x- and y- components of the initial velocity of the jet can be written as

While the components of the velocity of the wind are

So the components of the resultant velocity of the jet are

And the new speed is the magnitude of the resultant velocity:
