Hello!
Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.
Data:
Hooke represented mathematically his theory with the equation:
F = K * Δx
On what:
F (elastic force) = 2 N
K (elastic constant) = 4 N/cm
Δx (deformation or elongation of the elastic medium or distance from a spring) = ?
Solving:




simplify by 2


Answer:
B.) 1/2 cm
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I Hope this helps, greetings ... Dexteright02! =)
Answer:
u need to make sure that comparison is = to shapes and then find the shapes sizes and add them
Draw a right triangle so that its hypotenuse is 600 ft. The adjacent side is below the vertical, and it makes an angle of 75° with the hypotenuse.
Let h = height of the right triangle.
By definition,
sin75° = h/600
h = 600*sin75° = 579.555 = 580 ft (nearest ft)
Answer: 580 ft (nearest foot)
Answer:
1.137278672 m/s
+5.9 cm or -5.9 cm
Explanation:
A = Amplitude = 6.25 cm
m = Mass of object = 225 g
k = Spring constant = 74.5 N/m
Maximum speed is given by

The maximum speed of the object is 1.137278672 m/s
Velocity is at any instant is given by

The locations are +5.9 cm or -5.9 cm