Answer:
s = 589.3 m
Explanation:
Let the truck and car meet at a distance = s m
The truck is moving at constant velocity = v
so s= v * t ---------- (1)
car:
Vi = 0 m/s
a = 3.9 m/s²
s = Vi* t + 1/2 a t²
s= 0 * t + 1/2 a t²
s = 1/2 a t² ----------- (2)
compare equation (1) and equation (2)
s= v * t = 1/2 a t²
⇒ v * t = 1/2 a t²
⇒ t = 2 * v/ a
⇒ t = (2 * 33.9 )/ 3.9
⇒ t = 17. 38 s
Now
from equation (1)
s= v * t
s= 33.9 * 17.38
⇒ s = 589.3 m
In order to decrease the friction on the slide,
we could try some of these:
-- Install a drippy pipe across the top that keeps continuously
dripping olive oil on the top end of the slide. The oil oozes
down the slide and keeps the whole slide greased.
-- Hire a man to spread a coat of butter on the whole slide,
every 30 minutes.
-- Spray the whole slide with soapy sudsy water, every 30 minutes.
-- Drill a million holes in the slide,and pump high-pressure air
through the holes. Make the slide like an air hockey table.
-- Keep the slide very cold, and keep spraying it with a fine mist
of water. The water freezes, and a thin coating of ice stays on
the slide.
-- Ask a local auto mechanic to please, every time he changes
the oil in somebody's car, to keep all the old oil, and once a week
to bring his old oil to the park, to spread on the slide. If it keeps
the inside of a hot car engine slippery, it should do a great job
keeping a simple park slide slippery.
-- Keep a thousand pairs of teflon pants near the bottom of the ladder
at the beginning of the slide. Anybody who wants to slide faster can
borrow a set of teflon pants, put them on before he uses the slide, and
return them when he's ready to go home from the park.
Answer:
I_weight = M L²
this value is much larger and with it it is easier to restore balance.I
Explanation:
When man walks a tightrope, he carries a linear velocity, this velocity is related to the angular velocity by
v = w r
For man to maintain equilibrium needs the total moment to be zero
∑τ = I α
S τ = 0
The forces on the home are the weight of the masses, the weight of the man and the support on the rope, the latter two are zero taque the distance to the center of rotation is zero.
Therefore the moment of the masses and the open is the one that must be zero.
If the man carries only the bar, we could approximate it by two open one on each side of the axis of rotation formed by the free of the rope
I = ⅓ m L² / 4
As the length of half the length of the bar and the mass of the bar is small, this moment is small, therefore at the moment if there is some imbalance it is difficult to recover.
If, in addition to the opening, each of them carries a specific weight, the moment of inertia of this weight is
I_weight = M L²
this value is much larger and with it it is easier to restore balance.
