Answer:
The answer to the question is
The object would fall 57.625 m in the first 5 seconds
Explanation:
To solve the question, we note that
the height of fall = 490 ft = 149.352 m
Time to touch the ground = 7 seconds
We are required to find out how far the object falls in the first 5 seconds
We apply the relation
S = u·t + 0.5×g·t ² = We then have
149.352 = U×7+0.5*9.81*49 From where u = -13 m/s
Therefore to find how far it falls in the first 5 seconds, we have
-13*5 + 0.5*9.81*25 = 57.625 m
The force applied on the spring to stretch it is 13.2 N.
Hooke's law is a law of elasticity discovered by the English scientist Robert Hooke in 1660 that states that the displacement or size of a deformation is directly proportional to the deforming force or load for relatively small deformations of an object. When the load is removed under these conditions, the object returns to its original shape and size.
According to Hooke's law, F = k*e
where F is the force on the spring
k is force constant
and e is extension
F = (110)*(0.12)
F = 13.2 N
For more information on Hooke's law, visit :
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Answer:
the time it takes for one complete back and forth swing
Explanation:
the Mark's is showing you the time it swings back and forth
Greenhouse Gases, on relation to Earth's atmosphere.
Answer: just do the same thing, but the problems are different
Explanation: try you best
