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
Answer:
Light comes in different colors like radio, ultra violet, gamma-ray, etc, and they are invisible to the bare eye
Explanation:
Answer:
0.266 m
Explanation:
Assuming the lump of patty is 3 Kg then applying the principal of conservation of linear momentum,
P= mv where p is momentum, m is mass and v is the speed of an object. In this case
where sunscripts p and b represent putty and block respectively, c is common velocity.
Substituting the given values then
3*8=v(15+3)
V=24/18=1.33 m/s
The resultant kinetic energy is transferred to spring hence we apply the law of conservation of energy
where k is spring constant and x is the compression of spring. Substituting the given values then

225 = 1/2 (50) (v2)
225 = 25 (v2)
225/25 = v2
9 = v2
√9 = v
v = 3 m/s
Hi there!
We can use the rotational equivalent of Newton's Second Law:

Στ = Net Torque (Nm)
I = Moment of inertia (kgm²)
α = Angular acceleration (rad/sec²)
We can plug in the given values to solve.
