Answer:8 meters
Explanation:100cm is a meter
Answer:
lowest frequency = 535.93 Hz
distance between adjacent anti nodes is 4.25 cm
Explanation:
given data
length L = 32 cm = 0.32 m
to find out
frequency and distance between adjacent anti nodes
solution
we consider here speed of sound through air at room temperature 20 degree is approximately v = 343 m/s
so
lowest frequency will be =
..............1
put here value in equation 1
lowest frequency will be =
lowest frequency = 535.93 Hz
and
we have given highest frequency f = 4000Hz
so
wavelength =
..............2
put here value
wavelength =
wavelength = 0.08575 m
so distance =
..............3
distance =
distance = 0.0425 m
so distance between adjacent anti nodes is 4.25 cm
C
I took the tests earlier hope this helps
Answer:

Explanation:
From frequency of oscillation

Initially with the suspended string, the above equation is correct for the relation, hence

where k is force constant and m is the mass
When the spring is cut into half, by physics, the force constant will be doubled as they are inversely proportional

Employing f2/ f1, we have

Answer:
force of the breaks is 6650 N, direction opposite to direction of movement
Explanation: