Answer:
The distance by which the spring stretches is 1.48 cm.
Explanation:
Given that,
An object on a vertical spring oscillates up and down in simple harmonic motion with an angular frequency of 25.7 rad/s,
We know that angular frequency in SHM is given by :
When the object is allowed to hang stationary from it, the force due to spring is balanced by its weight. Such that :
From equation (1) :
So, the distance by which the spring stretches is 1.48 cm.
Answer:
160 ft
Explanation:
As the canon ball was shot horizontally, its initial vertical velocity is 0. The ball vertical motion is generated by the gravitational acceleration g = 32m/s2. We can calculate the time it takes for it to drop 64 ft
This is also the time it takes to travel horizontally at a constant rate of 80 ft/s if we ignore air resistance.
So the ball would land 160ft away from the wall on the ground
We know that the formula to of acceleration is given by,
Re-arrange to the Velocity,
(1)
We know as well for second Newton's law is,
We can re-arrange for the acceleration, so
(2)
Replacing (2) to (1),
Our values are,
Substituting the values,
We have two components of a net velocity, so we can calculate the direct force through,