Full Question
A 10 kilogram object suspended from the end of a vertically hanging spring stretches the spring 9.8 centimeters. At time t = 0, the resulting mass—spring system is disturbed from its rest state by the force F (t) = 20 cos(8t). The force F (t) is expressed in Newtons and is positive in the downward direction, and time is measured in seconds.
Determine the spring constant k.
Answer:
Spring Constant = 1000Nm^-1
Step-by-step explanation:
Given
Mass = 10kg
Spring Extension = 9.8cm
Spring Constant K = Force/Extension
Force = Mass * Acceleration
Force = 10 * 9.8 = 9.8N
Spring Extension = 9.8 cm = 9.8 /100 = 0.098 m
Spring constant = 98 / 0.098
Spring Constant = 1000Nm^-1
Answer:
press this link and u will get the answers
https://www.math.tamu.edu/~rosanna/141/WIR2_done.pdf
A+b=15----(1)
a*b=54----(2)
(1) a=15-b ----(3)
put(3) in(2)
(15-b)*b=54
15b-b^2=54
b^2-15b+54=0
(b-9)(b-6)=0
so b=9 or b=6
if b=9 then a=15-9=6
if b=6 then a=15-6=9
answer two numbers are 6 and 9
Answer:
Step-by-step explanation:
= 9 - 7x
3 times 2 -7 x + 3
6 - 7 x + 3
Answer:
18
Step-by-step explanation:
6 + 6 + 6 = 18
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