Answer:
The angular acceleration of the pencil<em> α = 17 rad·s⁻²</em>
Explanation:
Using Newton's second angular law or torque to find angular acceleration, we get the following expressions:
τ = I α (1)
W r = I α (2)
The weight is that the pencil has is,
sin 10 = r / (L/2)
r = L/2(sin(10))
The shape of the pencil can be approximated to be a cylinder that rotates on one end and therefore its moment of inertia will be:
I = 1/3 M L²
Thus,
mg(L / 2)sin(10) = (1/3 m L²)(α)
α(f) = 3/2(g) / Lsin(10)
α = 3/2(9.8) / 0.150sin(10)
<em> α = 17 rad·s⁻²</em>
Therefore, the angular acceleration of the pencil<em> </em>is<em> 17 rad·s⁻²</em>
A=(vf-vi)/t
a=(50-25)/10
a=2.5m/s^2
Answer:
0.09 x10^-10m
Explanation:
Using wavelength=( 12.27 A)/√V
= 12.27 x 10^-10/ √1.6x10^2
= 0.09x10^-10m
The output waveforms after passing through the transformer actually depend on the type of transformer used. It could either be a step-up transformer (steps voltage up), or a step-down transformer (steps voltage down). Both transformers have an output voltage in a form of a sine wave.
