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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! =)
Do you have a picture of the diagram that I could view?
Answer:
2.11 seconds
Explanation:
We use the kinematic equation for the velocity in a constantly accelerated motion under the acceleration of gravity (g):

Answer:
1.22gcm³
Explanation:
D = mass/ volume
Mass=1.840kg = 1,840g
1000g = 1kg
1.840kg= x(g)
X(g) = 1.840/1000
= 1840g
Volume = 0.0015m³= 1,500cm³
1m³= 1000,000cm³.
0.0015m³= x(cm³)
X(cm³) = 1000,000×0.0015
X(cm³)= 1500.
Since density is mass/volume, now impute your data's
D=m/v
D=1840/1500
1.22g/cm³