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:
Explanation:
- The expression for acceleration of the rolling body on an inclined plane is given as a = gsinФ/1 + k²/R²
- where Ф is the angle of inclination, R is the radius, k is the radius of gyration.
- The potential energy of the system is given as ; PE = mgh
- The potential energy will be constant for ring, cylinder, solid sphere, and hollow sphere.
- The total kinetic energy of the rolling body is ; KE = mv²/2 + Iw²/2
- Hence, the total kinetic energy of the ring, cylinder, solid sphere and hollow sphere will be constant.
2. The moment of inertia of the ring is given as ;
I = mR²
The moment of inertia of the ring is maximum and therefore reaches the bottom last.
Answer:
W = 1418.9 J = 1.418 KJ
Explanation:
In order to find the work done by the pull force applied by Karla, we need to can use the formula of work done. This formula tells us that work done on a body is the product of the distance covered by the object with the component of force applied in the direction of that displacement:
W = F.d
W = Fd Cosθ
where,
W = Work Done = ?
F = Force = 151 N
d = distance covered = 10 m
θ = Angle with horizontal = 20°
Therefore,
W = (151 N)(10 m) Cos 20°
<u>W = 1418.9 J = 1.418 KJ</u>
