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! =)
Using your periodic table if you look at it 3-11 are tansition metals so the horizontal Group Number will help if the group number has to digits just remove the one so if it were to be 13, the valence would be 3, if it were 14 the valence would be ,4 if it were 15, the valence would be 5, if it were 16 the valence would be 6, if it were 17 the valence would be 7 if it were group 18 the valence would be 8 so if anymore help needed to explain hit me up
I’m pretty sure
Part 1; C
Part 2; C
Not 100% sure tho :)
Explanation:
The aircraft is traveling north at 100 m/s.
The wind blows from the west (towards the east) at 25 m/s.
The two vectors form a right triangle. The magnitude of the resultant velocity can be found with Pythagorean theorem:
v² = vx² + vy²
v² = (25 m/s)² + (100 m/s)²
v = 103 m/s
The direction can be found with trigonometry:
θ = atan(vy / vx)
θ = atan(100 / 25)
θ = 76.0°
The resultant velocity is 103 m/s at 76.0° north of east.
A: A resource that will always be there, can be replenished by the biogeochemical cycles. B: Can regenerate if they are alive or can be replenished by biochemical cycles if they are non living.
