U can always just do the classic roller coaster going up an incline and create some sort of story from that.
Answer: perpendicular to it oscillations.
Explanation: A transverse wave is a wave whose oscillations is perpendicular to the direction of the wave.
By perpendicular, we mean that the wave is oscillating on the vertical axis (y) of a Cartesian plane and the vibration is along the horizontal axis (x) of the plane.
Examples of transverse waves includes wave in a string, water wave and light.
Let us take a wave in a string for example, you tie one end of a string to a fixed point and the other end is free with you holding it.
If you move the rope vertically ( that's up and down) you will notice a kind of wave traveling away from you ( horizontally) to the fixed point.
Since the oscillations is perpendicular to the direction of wave, it is a transverse wave
Answer:
a. 16 s b. -1.866 kJ
Explanation:
a. Since the initial rotational speed ω₀= 3313 rev/min = 3313/60 × 2π rad/s = 346.94 rad/s. Its rotational speed becomes ω₁ = 0.75ω₀ in time t = 4 s.
We find it rotational acceleration using α = (ω₁ - ω₀)/t = (0.75ω₀ - ω₀)/t = ω₀(0.75 - 1)/t = -0.25ω₀/t = (-0.25 × 346.94 rad/s)/4 s = -21.68 rad/s².
Since the turntable stops at ω = 0, the time it takes to stop is gotten from
ω = ω₀ + αt and t = (ω - ω₀)/α = (0 - 346.94 rad/s)/-21.68 rad/s² = (-346.94/-21.68) s = 16 s.
So it takes the turntable 16 s to stop.
b. The workdone by the turntable to stop W equals its rotational kinetic energy change.
So, W = 1/2Iω² - 1/2Iω₀² = 1/2 × 0.031 kgm² × 0² - 1/2 × 0.031 kgm² × (346.94 rad/s)² = 0 - 1865.7 J = -1865.7 J = -1.8657 kJ ≅ -1.866 kJ
Use stronger magnets
increase current
push magnets closer to coil
adding more sets of coils
Answer:
The elastic potential energy of the spring change during this process is 21.6 J.
Explanation:
Given that,
Spring constant of the spring, 
It extends 6 cm away from its equilibrium position.
We need to find the elastic potential energy of the spring change during this process. The elastic potential energy of the spring is given by the formula as follows :
So, the elastic potential energy of the spring change during this process is 21.6 J.
