Answer:
more than 500 n i think the answer will
Answer: B
Explanation:
Given that an object of mass 2 kg starts from rest and is allowed to slide down a frictionless incline so that its height changes by 20 m.
The parameters given from the question are:
Mass M = 2kg
Height h = 20m
Let g = 9.8m/s^2
At the bottom of the incline plane, the object will experience maximum kinetic energy.
From conservative of energy, maximum K.K.E = maximum P.E
Maximum P.E = mgh
Maximum P.E = 2 × 9.8 × 20 = 392 J
But
K.E = 1/2mv^2
Substitute the values of energy and mass into the formula
392 = 1/2 × 2 × V^2
V^2 = 392
V = sqrt( 392 )
V = 19.8 m/s
V = 20 m/s approximately
Answer:
See below
Explanation:
F = ma
F = 12 * 9 = 108 N
108 N needed <u> add 30 N more east </u>
Answer:
75 rad/s
Explanation:
The angular acceleration is the time rate of change of angular velocity. It is given by the formula:
α(t) = d/dt[ω(t)]
Hence: ω(t) = ∫a(t) dt
Also, angular velocity is the time rate of change of displacement. It is given by:
ω(t) = d/dt[θ(t)]
θ(t) = ∫w(t) dt
θ(t) = ∫∫α(t) dtdt
Given that: α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴. Hence:
θ(t) = ∫∫α(t) dtdt
θ(t) = ∫∫6t² dtdt =∫[∫6t² dt]dt
θ(t) = ∫[2t³]dt = t⁴/2 rad
θ(t) = t⁴/2 rad
At θ(t) = 10 rev = (10 * 2π) rad = 20π rad, we can find t:
20π = t⁴/2
40π = t⁴
t = ⁴√40π
t = 3.348 s
ω(t) = ∫α(t) dt = ∫6t² dt = 2t³
ω(t) = 2t³
ω(3.348) = 2(3.348)³ = 75 rad/s