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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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A 0.83-kg block is hung from and stretches a spring that is attached to the ceiling. A second block is attached to the first one
vodomira [7]

Answer:

1.66 kg

Explanation:

Given that a 0.83-kg block is hung from and stretches a spring that is attached to the ceiling.

From Hook's law

F = Ke

But F = mg

Substitute mg for force in the Hook's law

Mg = ke

0.83 × 9.8 = ke

Make K the subject of formula

8.134 = Ke

K = 8.134 /e

Given that a second block is attached to the first one, and the amount that the spring stretches from its unstretched length triples.

That is

(0.83 + M) × 9.8 = K (3e)

Substitutes K into the above equation

(0.83 + M) × 9.8 = 8.134 / e (3e)

The e will cancel out

(0.83 + M) × 9.8 = 24.402

0.83 + M = 24.402/9.8

0.83 + M = 2.49

M = 2.49 - 0.83

M = 1.66 kg

Therefore, the mass of the second block is 1.66kg

6 0
3 years ago
g According to the Third Law, the action and reaction forces are exactly equal in magnitude and in opposite directions. So when
torisob [31]

Answer:

One way to look at this is to consider the forces acting on any point in a string.

For a very small portion of string F = M a must still hold. As M approaches zero the small portion of string would have to approach infinite acceleration if the net force on that portion of string were not zero.

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Can the momentum of an object change
OlgaM077 [116]
The answer for this is yes.
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Answer:

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uyui

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