Answer:
0.005 m
Explanation:
length of steel (L°) = 12 m
initial temperature (T) = 16 degrees
expected temperature (T') = 50 degrees
We can find how large the gaps should be if the track is not to buckle when the temperature is as high as 50 degrees from the formula below
ΔL = ∝L°ΔT where
ΔL = x 12 x (50-16) = 0.005 m
The complete question is as follows:
A 0.150-kg toy is undergoing SHM on the end of a horizontal spring with force constant k = 300 N/m. When the toy is 0.0120 m from its equilibrium position, it is observed to have a speed of 0.400 m/s.
Answer:
The correct answer is 0.034 J.
Explanation:
Given :
mass of the toy is m = 0.15 kg.
The force constant of restoring force k = 300 Nm⁻¹
When the position of the toy from the equilibrium is x = 0.012m, then the
speed of the toy vx = 0.4m s
The total mechanical energy in SHM is given by
E= 1/2 (mv²+ kx²) = 1/2 kA²
(here, m = mass of the object, vx = velocity, k = force constant
of restoring force, and A = amplitude of SHM.)
Hence by substituting the numerical values in equation 1, we get
= 0.034 J
Thus, the correct answer is 0.034 J.