Answer:
d. zero
Explanation:
Constant velocity means the acceleration is zero. In this case the velocity does not change,
hope this helps you
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<span>Answer:
So this involves right triangles. The height is always 100. Let the horizontal be x and the length of string be z.
So we have x2 + 1002 = z2. Now take its derivative in terms of time to get
2x(dx/dt) = 2z(dz/dt)
So at your specific moment z = 200, x = 100âš3 and dx/dt = +8
substituting, that makes dz/dt = 800âš3 / 200 or 4âš3.
Part 2
sin a = 100/z = 100 z-1 . Now take the derivative in terms of t to get
cos a (da./dt) = -100/ z2 (dz/dt)
So we know z = 200, which makes this a 30-60-90 triangle, therefore a=30 degrees or π/6 radians.
Substitute to get
cos (Ď€/6)(da/dt) = (-100/ 40000)(4âš3)
âš3 / 2 (da/dt) = -âš3 / 100
da/dt = -1/50 radians</span>
Answer:
a)- 1.799 rad/sec²
b)- 17.6 x 10ˉ³Nm
Explanation:
ω₀ = 720 rev/min x (1 min/60 sec) x (2π rad / 1 rev) = 24π rad/s
a) Assuming a constant angular acceleration, the formula will be
α = (ωf -ω₀) / t
As final state of the grindstone is at rest, so ωf =0
⇒ α = (0-24π) / 41.9 = - 1.799 rad/sec²
b)Moment of inertia I for a disk about its central axis
I = ½mr²
where m=2kg and radius 'r'= 0.099m
I = ½(2)(0.099²)
I = 9.8 x 10ˉ³ kgm²
Next is to determine the frictional torque exerted on the grindstone, that caused it to stop, applying the rotational equivalent of the Newton's 2nd law:
τ = I α =>(9.8 x 10ˉ³)(- 1.799)
τ = - 17.6 x 10ˉ³Nm
(The negative sign indicates that the frictional torque opposes to the rotation of the grindstone).
