The velocity of the package after it has fallen for 3.0 s is 29.4 m/s
From the question,
A small package is dropped from the Golden Gate Bridge.
This means the initial velocity of the package is 0 m/s.
We are to calculate the velocity of the package after it has fallen for 3.0 s.
From one of the equations of kinematics for objects falling freely,
We have that,
v = u + gt
Where
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity
and t is time
To calculate the velocity of the package after it has fallen for 3.0 s
That means, we will determine the value of v, at time t = 3.0 s
The parameters are
u = 0 m/s
g = 9.8 m/s²
t = 3.0 s
Putting these values into the equation
v = u + gt
We get
v = 0 + (9.8×3.0)
v = 0 + 29.4
v = 29.4 m/s
Hence, the velocity of the package after it has fallen for 3.0 s is 29.4 m/s
Learn more here: brainly.com/question/13327816
Answer:
Explanation:
The collision is inelastic and can be described by the Principle of Momentum Conservation:
The speed after the collision is:
If you will be rinsing your regulator after removing it from the cylinder, you must make sure that the dust cap is firmly in place.
<h3>What is Dust cap?</h3>
A dust cap is a gently curved dome mounted either in concave or convex orientation over the central hole of most loudspeaker diaphragms.
Thus, if you will be rinsing your regulator after removing it from the cylinder, you must make sure that the dust cap is firmly in place.
Learn more about dust cap here: brainly.com/question/10723016
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Answer:
the answer choices are at the bottom right?
Explanation:
