Answer:
The acceleration of the collar is 10 m/s²
Explanation:
Given;
mass of the collar, m = 1 kg
applied force on the bar, F = 10 N
The acceleration of the collar can be calculated by applying Newton's second law of motion;
F = ma
where;
F is the applied force
m is mass of the object
a is the acceleration
a = F / m
a = 10 / 1
a = 10 m/s²
Therefore, the acceleration of the collar is 10 m/s²
-- Although it's not explicitly stated in the question,we have to assume that
the surface is frictionless. I guess that's what "smooth" means.
-- The total mass of both blocks is (1.5 + 0.93) = 2.43 kg. Since they're
connected to each other (by the string), 2.43 kg is the mass you're pulling.
-- Your force is 6.4 N.
Acceleration = (force)/(mass) = 6.4/2.43 m/s²<em>
</em> That's about <em>2.634 m/s²</em> <em>
</em>(I'm going to keep the fraction form handy, because the acceleration has to be
used for the next part of the question, so we'll need it as accurate as possible.)
-- Both blocks accelerate at the same rate. So the force on the rear block (m₂) is
Force = (mass) x (acceleration) = (0.93) x (6.4/2.43) = <em>2.45 N</em>.
That's the force that's accelerating the little block, so that must be the tension
in the string.
Explanation:
In total, the length is measured from the tip of the bow in a linear fashion to the stern of the formation of delight including any back-deck extensions. The measurement involves bow sprits; rudders; detachable engines and engine sections; handles; and various fittings and connections.
Importance in calculating a boat's length:
it affects the transportation costs (the longer the length, the higher the cost).
The pontoon's length counts as you find out how much rope you need to wrestle.
The cost of vessel settlement on marinas depends in part on the pontoon length. As more area is consumed by a more drawn pontoon, the docking charges are higher.
Transportation guidelines will probably not allow pontoons past a specific length on specific occasions of the day.