Answer:
The spring constant of this spring is 200 N/m.
Explanation:
Given:
Original unstretched length of the spring (x₀) = 10 cm =0.10 m [1 cm =0.01 m]
Stretched length of the spring (x₁) = 18 cm = 0.18 cm
Force acting on the spring (F) = 16 N
Spring constant of the spring (k) = ?
First let us find the change in length of the spring or the elongation caused in the spring due to the applied force.
So, Change in length = Final length - Initial length

Now, restoring force acting on the spring is directly related to its elongation or compression as:

Rewriting in terms of 'k', we get:

Now, plug in the given values and solve for 'k'. This gives,

Therefore, the spring constant of this spring is 200 N/m.
Answer: it should be cut with a chainsaw
Explanation:
Answer:
53.895 m.
Explanation:
Using the equation of motion,
v² = u² + 2as .............. Equation 1
Where v = final velocity of the swan, u = initial velocity of the swan, a = acceleration of the swan, s = distance covered by the swan.
make s the subject of the equation,
s = (v² - u²)/2a----------- Equation 2
Given: v = 6.4 m/s, u = 0 m/s ( from rest) a = 0.380 m/s².
Substitute into equation 2
s = (6.4²-0²)/(2×0.380)
s = 40.96/0.76
s = 53.895 m.
Hence the swan will travel 53.895 m before becoming airborne.