Answer:
ω = 3.61 rad/sec
Explanation:
Firstly, we should know that the bug will not slip if friction can provide sufficient opposing force.
μmg = mv^2/r = mω^2r
Thus;
μg = ω^2r
ω^2 = μg/r
ω = √(μg/r)
ω = √(0.321 * 9.8)/0.241
ω = √(13.05)
= 3.61 rad/sec
Question is from B to C
Answer: (b) 1.5m/s
x1=3m, x2=9m
t1=1s, t2=5s
Displacement, ∆x=(9-3)m=6m
Time elapsed, ∆t=(5-1)s=4s
So average velocity v =∆x/∆t=6/4=1.5m/s
Answer:
<h2>10 kg.m/s</h2>
Explanation:
The momentum of an object can be found by using the formula
momentum = mass × velocity
From the question we have
momentum = 20 × 0.5
We have the final answer as
<h3>10 kg.m/s</h3>
Hope this helps you
Answer:
hmax = 1/2 · v²/g
Explanation:
Hi there!
Due to the conservation of energy and since there is no dissipative force (like friction) all the kinetic energy (KE) of the ball has to be converted into gravitational potential energy (PE) when the ball comes to stop.
KE = PE
Where KE is the initial kinetic energy and PE is the final potential energy.
The kinetic energy of the ball is calculated as follows:
KE = 1/2 · m · v²
Where:
m = mass of the ball
v = velocity.
The potential energy is calculated as follows:
PE = m · g · h
Where:
m = mass of the ball.
g = acceleration due to gravity (known value: 9.81 m/s²).
h = height.
At the maximum height, the potential energy is equal to the initial kinetic energy because the energy is conserved, i.e, all the kinetic energy was converted into potential energy (there was no energy dissipation as heat because there was no friction). Then:
PE = KE
m · g · hmax = 1/2 · m · v²
Solving for hmax:
hmax = 1/2 · v² / g
Answer:
The answer is A. C and O..