<h3>Answer</h3>
m/s^2 (meter per sec square)
Explanation:
acc = change in velocity/time
= distance/time
----------------
time
= m/s
------
s
=m/s^2
Answer:
a) Δx = 49.23 mi
, b) Δx = 5.77 mi
Explanation:
As we have an acceleration function we must use the definition of kinematics
a = dv / dt
∫dv = ∫ a dt
we integrate and evaluators
v - vo = ∫ (-1280 (1 + 8t)⁻³ dt = -1280 ∫ (1+ 8t)⁻³ dt
We change variables
1+ 8t = u
8 dt = du
v - v₀ = -1280 ∫ u⁻³ du / 8
v -v₀ = -1280 / 8 (-u⁻²/2)
v - v₀ = 80 (1+ 8t)⁻²
We evaluate between the initial t = 0 v₀ = 80 and the final instant t and v
v- 80 = 80 [(1 + 8t)⁻² - 1]
v = 80 (1+ 8t)⁻²
We repeat the process for defining speed is
v = dx / dt
dx = vdt
x-x₀ = 80 ∫ (1-8t)⁻² dt
x-x₀ = 80 ∫ u⁻² dt / 8
x-x₀ = 80 (-1 / u)
x-x₀ = -80 (1 / (1 + 8t))
We evaluate for t = 0 and x₀ and the upper point t and x
x -x₀ = -80 [1 / (1 + 8t) - 1]
We already have the function of time displacement
a) let's calculate the position at the two points and be
t = 0 h
x = x₀
t = 0.2 h
x-x₀ = -80 [1 / (1 +8 02) -1]
x-x₀ = 49.23
displacement is
Δx = x (0.2) - x (0)
Δx = 49.23 mi
b) in the interval t = 0.2 h at t = 0.4 h
t = 0.4h
x- x₀ = -80 [1 / (1+ 8 0.4) -1]
x-x₀ = 55 mi
Δx = x (0.4) - x (0.2)
Δx = 55 - 49.23
Δx = 5.77 mi
Answer:
When a an object is been rotated its resistance capacity to that rotational force is know as rotational inertia and this mathematically given as
Where m is the mass
r is the rotation radius
For the spinning of the lamp as a baton to work the location of the center of mass of the floor lamp needs to be located
This is more likely to be located closer to base of the lamp as compared to the top, so success of spinning a floor lamp like a baton is highly likely if the lamp is grabbed closer to the base because that is where the position of its center of mass is likely to be.
Explanation:
The force on the tool is entirely in the negative-y direction.
So no work is done during any moves in the x-direction.
The work will be completely defined by
(Force) x (distance in the y-direction),
and it won't matter what route the tool follows to get anywhere.
Only the initial and final y-coordinates matter.
We know that F = - 2.85 y². (I have no idea what that ' j ' is doing there.)
Remember that 'F' is pointing down.
From y=0 to y=2.40 is a distance of 2.40 upward.
Sadly, since the force is not linear over the distance, I don't think
we can use the usual formula for Work = (force) x (distance).
I think instead we'll need to integrate the force over the distance,
and I can't wait to see whether I still know how to do that.
Work = integral of (F·dy) evaluated from 0 to 2.40
= integral of (-2.85 y² dy) evaluated from 0 to 2.40
= (-2.85) · integral of (y² dy) evaluated from 0 to 2.40 .
Now, integral of (y² dy) = 1/3 y³ .
Evaluated from 0 to 2.40 , it's (1/3 · 2.40³) - (1/3 · 0³)
= 1/3 · 13.824 = 4.608 .
And the work = (-2.85) · the integral
= (-2.85) · (4.608)
= - 13.133 .
-- There are no units in the question (except for that mysterious ' j ' after the 'F',
which totally doesn't make any sense at all).
If the ' F ' is newtons and the 2.40 is meters, then the -13.133 is joules.
-- The work done by the force is negative, because the force points
DOWN but we lifted the tool UP to 2.40. Somebody had to provide
13.133 of positive work to lift the tool up against the force, and the force
itself did 13.133 of negative work to 'allow' the tool to move up.
-- It doesn't matter whether the tool goes there along the line x=y , or
by some other route. WHATEVER the route is, the work done by ' F '
is going to total up to be -13.133 joules at the end of the day.
As I hinted earlier, the last time I actually studied integration was in 1972,
and I haven't really used it too much since then. But that's my answer
and I'm stickin to it. If I'm wrong, then I'm wrong, and I hope somebody
will show me where I'm wrong.
