Answer:
true , I searched and got u the answer
Answer:
Later high school years and freshman year of college
Explanation:
The transition from high school to college is an important developmental milestone that holds the potential for personal growth and behavioral change. A cohort of 2,025 students was recruited during the summer before they matriculated into college and completed Internet-based surveys about their participation in a variety of behavioral risks during the last three months of high school and throughout the first year of college. Alcohol use, marijuana use, and sex with multiple partners increased during the transition from high school to college, whereas driving after drinking, aggression, and property crimes decreased. Those from rural high schools and those who elected to live in private dormitories in college were at highest risk for heavy drinking and driving after drinking.
Well, first of all, one who is sufficiently educated to deal with solving
this exercise is also sufficiently well informed to know that a weighing
machine, or "scale", should not be calibrated in units of "kg" ... a unit
of mass, not force. We know that the man's mass doesn't change,
and the spectre of a readout in kg that is oscillating is totally bogus.
If the mass of the man standing on the weighing machine is 60kg, then
on level, dry land on Earth, or on the deck of a ship in calm seas on Earth,
the weighing machine will display his weight as 588 newtons or as
132.3 pounds. That's also the reading as the deck of the ship executes
simple harmonic motion, at the points where the vertical acceleration is zero.
If the deck of the ship is bobbing vertically in simple harmonic motion with
amplitude of M and period of 15 sec, then its vertical position is
y(t) = y₀ + M sin(2π t/15) .
The vertical speed of the deck is y'(t) = M (2π/15) cos(2π t/15)
and its vertical acceleration is y''(t) = - (2πM/15) (2π/15) sin(2π t/15)
= - (4 π² M / 15²) sin(2π t/15)
= - 0.1755 M sin(2π t/15) .
There's the important number ... the 0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.
The net vertical force on the intrepid sailor ... the guy standing on the
bathroom scale out on the deck of the ship that's "bobbing" on the
high seas ... is (the force of gravity) + (the force causing him to 'bob'
harmonically with peak acceleration of 0.1755 x amplitude).
At the instant of peak acceleration, the weighing machine thinks that
the load upon it is a mass of 65kg, when in reality it's only 60kg.
The weight of 60kg = 588 newtons.
The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
in order to accelerate him faster than gravity.
Now I'm going to wave my hands in the air a bit:
Apparent weight = (apparent mass) x (real acceleration of gravity)
(Apparent mass) = (65/60) = 1.08333 x real mass.
Apparent 'gravity' = 1.08333 x real acceleration of gravity.
The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
the bobbing of the deck.
0.08333 G = 0.1755 M
The 'M' is what we need to find.
Divide each side by 0.1755 : M = (0.08333 / 0.1755) G
'G' = 9.0 m/s²
M = (0.08333 / 0.1755) (9.8) = 4.65 meters .
That result fills me with an overwhelming sense of no-confidence.
But I'm in my office, supposedly working, so I must leave it to others
to analyze my work and point out its many flaws.
In any case, my conscience is clear ... I do feel that I've put in a good
5-points-worth of work on this problem, even if the answer is wrong .
Answer:
a) 
, b) 
Explanation:
a) Let consider two equations of equilibrium, the first parallel to ski slope and the second perpendicular to that. The equations are, respectively:
The force on the skier is:
b) The equations of equilibrium are the following:
The force on the skier is:
We calculate the coordinates at t₁ = 9 min and t₂ = 10 min, since the 10th minute is between t₁ and t₂.
As it leaves from rest, it means that the initial speed is zero
t₁=9 min=540 s
t₂=10 min=600 s
x₁=at₁²/2=8*540²/2=4*291600=1166400 m
x₂=at₂²/2=8*600²/2=4*360000=1440000 m
Δx=x₂-x₁=1440000-1166400=273600 m represents the distance traveled by the car in the 10th minute of travel
