An ornament of mass 42.0 g is attached to a vertical ideal spring with a force constant (spring constant) of 33.9 N/m. The ornam
1 answer:
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Answer:
Explanation:
<u>Instant Velocity and Acceleration </u>
Give the position of an object as a function of time y(x), the instant velocity can be obtained by
Where y'(x) is the first derivative of y respect to time x. The instant acceleration is given by
We are given the function for y
Note we have changed the last term to be quadratic, so the question has more sense.
The velocity is
And the acceleration is
Answer:
987 joules, 3.01s
Explanation:
(A)
from the attached diagram
net force, Fnet, pulling the crate up the ramp is given by
Fnet = FcosФ - WsinФ - Fr
where FcosФ is the component of horizontal force 290N resolved parallel to the plane
WsinФ = mgsinФ = component of the weight of the crate resolved parallel to the plane
Fr = constant opposing frictional force
Fnet = 290cos34⁰ - 20 × 9.8 × sin34° - 65
Fnet = 240.421 - 109.602 - 65
Fnet = 65.82N
Work done on the crate up the ramp, W, is given by
W = Fnet × d (distance up the plane)
W = 65.819 × 15
W = 987.285 joules
W = 987 joules (to 3 significant Figures)
(B)
to calculate the time of travel up the ramp
we use the equation of motion
where s = distance up the plane, 15m
u = Initial velocity of the crate, which is 0 for a body that is initially at rest
a = acceleration up the plane, given by
where m = mass of the crate, 20 kg
now,
from,
15 = 0 + 1.645
15 = 1.645
t = 3.01s (to 3 sig fig)
Answer:
c. Vessel side holes
Explanation:
- The "Poiseuille formula" which is given by
describes the volumetric flow rate (
) through tubular sections.
- Here,
represent the injection pressure difference, the length of the section, the radius of the section and the viscosity index of the fluid that flows through the section respectively.
- With this, one can confirm that all the factors except the vessel side holes affect the flow rate.
- Side holes, however, are a factor that could give a measure of how much volume would flow to a particular location. In such a situation the flow rate remains unchanged and one location would receive a lower volume (not the whole) as some volume would spill out at the side holes.
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