6. Drop to one quarter of its original value
Try to have a calm morning before camp. A good night’s sleep and a good breakfast. Make sure to be cautious, follow all the rules for certain areas ( some maybe restricted ). Take lots of photos doing wacky stuff! Learn but have fun learning
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:
Explanation:
Given that,
Mass of ball m = 2kg
Ball traveling a radius of r1= 1m.
Speed of ball is Vb = 2m/s
Attached cord pulled down at a speed of Vr = 0.5m/s
Final speed V = 4m/s
Let find the transverse component of the final speed using
V² = Vr²+ Vθ²
4² = 0.5² + Vθ²
Vθ² = 4²—0.5²
Vθ² = 15.75
Vθ =√15.75
Vθ = 3.97 m/s.
Using the conservation of angular momentum,
(HA)1 = (HA)2
Mb • Vb • r1 = Mb • Vθ • r2
Mb cancels out
Vb • r1 = Vθ • r2
2 × 1 = 3.97 × r2
r2 = 2/3.97
r2 = 0.504m
The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m
The required time,
Using equation of motion
V = ∆r/t
Then,
t = ∆r/Vr
t = (r1—r2) / Vr
t = (1—0.504) / 0.5
t = 0.496/0.5
t = 0.992 second
