Answer:
the net toque is τ=8.03* 10⁻⁴ N*m
Explanation:
Assuming the disk has constant density ρ, the moment of inertia I of is
I = ∫r² dm
since m = ρ*V = ρπR² h , then dm= 2ρπh r dr
thus
I = ∫r²dm = ∫r²2ρπh r dr =2ρπh ∫r³ dr = 2ρπh (R⁴/4- 0⁴/4)= ρπhR⁴ /2= mR²/2
replacing values
I = mR²/2= 0.017 kg * (0.06 m)²/2 = 3.06 *10⁻⁵ kg*m²
from Newton's second law applied to rotational motion
τ= Iα , where τ=net torque and α= angular acceleration
since the angular velocity ω is related with the angular acceleration through
ω= ωo + α*t → α =(ω-ωo)/t = (21 rad/s-0)/0.8 s = 26.25 rad/s²
therefore
τ= Iα= 3.06 *10⁻⁵ kg*m²*26.25 rad/s² = 8.03* 10⁻⁴ N*m
Answer:
A) The particle will accelerate in the direction of point C.
Explanation:
As we know that
potential at points A, B,C and D as V_A, V_B, V_C, V_D and it is clear from the question that
V_A>V_B>V_C
And we know that flow is always from higher to lower potential (for positive charge due to positive potential energy).
So the charge will accelerate from B toward C.
Hence, the correct option is A.
A = 4\pi r^2
A = 4\pi (2\mu m /2)^2 (10^{-6}m/1\mu m)^2 (1mm/10{-3})^2
A = 1.33*!0^{-5}MM^2
Explanation:
For each object, the initial potential energy is converted to rotational energy and translational energy:
PE = RE + KE
mgh = ½ Iω² + ½ mv²
For the marble (a solid sphere), I = ⅖ mr².
For the basketball (a hollow sphere), I = ⅔ mr².
For the manhole cover (a solid cylinder), I = ½ mr².
For the wedding ring (a hollow cylinder), I = mr².
If we say k is the coefficient in each case:
mgh = ½ (kmr²) ω² + ½ mv²
For rolling without slipping, ωr = v:
mgh = ½ kmv² + ½ mv²
gh = ½ kv² + ½ v²
2gh = (k + 1) v²
v² = 2gh / (k + 1)
The smaller the value of k, the higher the velocity. Therefore:
marble > manhole cover > basketball > wedding ring
