<h3><u>Answer</u>;</h3>
≈ 5 Kgm²/sec
<h3><u>Explanation</u>;</h3>
Angular momentum is given by the formula
L = Iω, where I is the moment of inertia and ω is the angular speed.
I = mr², where m is the mass and r is the radius
= 0.65 × 0.7²
= 0.3185
Angular speed, ω = v/r
= (2 × 3.142 × r × 2.5) r
= 15.71
Therefore;
Angular momentum = Iω
= 0.3185 × 15.71
= 5.003635
<u>≈ 5 Kgm²/sec</u>
Answer:
the average force 11226 N
Explanation:
Let's analyze the problem we are asked for the average force, during the crash, we can find this from the impulse-momentum equation, but this equation needs the speeds and times of the crash that we could look for by kinematics.
Let's start looking for the stack speeds, it has a free fall, from rest (Vo=0)
Vf² = Vo² - 2gY
Vf² = 0 - 2 9.8 7.69 = 150.7
Vf = 12.3 m / s
This is the speed that the battery likes when it touches the beam. They also give us the distance it travels before stopping, let's calculate the time
Vf = Vo - g t
0 = Vo - g t
t = Vo / g
t = 12.3 / 9.8
t = 1.26 s
This is the time to stop
Now let's use the equation that relates the impulse to the amount of movement
I = Δp
F t = pf-po
The amount of final movement is zero because the system stops
F = - po / t
F = - mv / t
F = - 1150 12.3 / 1.26
F = -11226 N
This is the average force exerted by the stack on the vean
Answer:
n = 5 approx
Explanation:
If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back
= e ( coefficient of restitution ) = 
and
h₁ is height up-to which the ball bounces back after first bounce.
From the two equations we can write that
So on
= .00396
Taking log on both sides
- n / 2 = log .00396
n / 2 = 2.4
n = 5 approx
By putting this special transportation plastic on the bottom of the sled, because the transportation plastic is slick. that is what the transport bins slide around on. (p.s) the plastic is really expensive!
