1.5 m/s is the velocity.
9.3 m is the length of aisle, over which Distance will be covered.
Time is demanded in which the child will move the cart over the aisle with 1.5 m/s.
v=S/t
and,
t=S/v
Put values,
t=9.3/1.5=6.2 s
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
I am going to say
C. Energy contained in the nucleus of an atom
