Missing question: "What is the spring's constant?"
Solution:
The object of mass m=6.89 kg exerts a force on the spring equal to its weight:
When the object is attached to the spring, the displacement of the spring with respect to its equilibrium position is
And by using Hook's law, we can find the constant of the spring:
Answer:
The answer to the questions is;
In terms of standing waves, the listener moves from a location with high amplitude to one with lower amplitude or vibration (anti-node to node)
The distance 4.1 cm is equivalent to λ/4
Explanation:
For standing waves we have is a stationary wave comprising of two opposite direction moving waves that have equal amplitude and frequency, resulting in the superimposition of the waves. As such certain points are fixed along the wave path that is the peaks amplitude of the wave oscillation is constant at a particular point. A node occurring at a point and an anti-node occurring at another fixed point
When the listener moves 4.1 cm he or she has left the anti-node to the node hence the faintness of the sound
The distance from the node to the anti-node is 1/4 wavelength, or 1/4×λ
Therefore 4.1 cm is λ/4
Answer:
Acceleration = 9 × 10^5 m/s^2 ( deceleration )
Explanation:
From the first equation of motion:
V = u + at
15000 = 30000 + 60a
a = ( 15000-30000)/60
a = 9 × 10^5 m/s^2
