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grin007 [14]
3 years ago
11

A hockey puck is set in motion across a frozen pond. If ice friction and air resistance are neglected, the forcerequired to keep

the puck sliding at constant velocity isA) the weight of the puck divided by the mass of the puck.B) the mass of the puck multiplied by 9.8 meters per second per second.~qual to the weight of the puck.'i.£l/ero newtons.E) none of these.
Physics
1 answer:
yulyashka [42]3 years ago
7 0

Answer:

Zero Newtons

Explanation:

Newton's second law of motion states that the net force applied to an object is equal to the product between the object's mass and its acceleration:

F=ma

In this case, we want the hockey puck to slide at constant velocity - constant velocity means zero acceleration:

a = 0

And so this means also that the net force is zero:

F = 0

However, the problem says that ice friction and air resistance are negligible - so there are no forces acting on the hockey puck. This means that the puck will continue its motion at constant velocity if we don't apply any other force on it.

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An important dimensionless parameter in certain types of fluid flow problems is the Froude number defined as V/√g.l where V is a
prohojiy [21]

Answer:

1.24611

Explanation:

V = Velocity = 10 ft/s

L = Length = 2 ft

g = Acceleration due to gravity = 32.2 ft/s²

Froude number is given by

Fr=\dfrac{V}{\sqrt{gL}}\\\Rightarrow Fr=\dfrac{10}{\sqrt{32.2\times 2}}\\\Rightarrow Fr=1.24611

Converting to SI units

10\ ft/s=10\times \dfrac{1}{3.281}

32.2\ ft/s^2=32.2\times \dfrac{1}{3.281}

2\ ft=2\times \dfrac{1}{3.281}

Fr=\dfrac{V}{\sqrt{gL}}\\\Rightarrow Fr=\dfrac{10\times \dfrac{1}{3.281}}{\sqrt{32.2\times \dfrac{1}{3.281}\times 2\times \dfrac{1}{3.281}}}\\\Rightarrow Fr=1.24611

The Froude number is 1.24611

The Froude number is equal. The Froude number is dimensionless as the units cancel each other. In order for this to happen the units used need to be consitent either imperial or SI.

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