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:
a) 1.22 s
b) 9.089 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
Time taken by the ball to reach the highest point is 1.22 seconds
The maximum height the ball will reach above the ground is 1.75+7.339 = 9.089 m
u might be right
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Answer:
0.48 m
Explanation:
I'm assuming that this takes place in an ideal situation, where we neglect a host of factors such as friction, weight of the spring and others
If the mass is hanging from equilibrium at 0.42 m above the floor, from the question, and it is then pulled 0.06 m below that particular position. This pulling is a means of adding more energy into the spring, when it is released, the weight compresses the spring and equals its distance (i.e, 0.06 m) above the height.
0.42 m + 0.06 m = 0.48 m
At the highest point thus, the height is 0.48 m above the ground.
