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LUCKY_DIMON [66]
3 years ago
5

in three situations, a briefly applied horizontal force changes the velocity of a hockey puck that slides over frictionless ice.

the overhead views of the figure indicate, for each situation, the puck’s initial speed vi, its final speed vf, and the directions of the corresponding velocity vectors. rank the situations according to the work done on the puck by the applied force, most positive first and most negative last.
Physics
1 answer:
victus00 [196]3 years ago
4 0

The work done on the Puck by the applied force from the most positive to the most negative is c, b, a respectively.

According to Newton's second law of motion, the force applied to an object is directly proportional to the product of mass and acceleration of the object.

F = ma

F= \frac{mv}{t}

The force applied to an object increases with increases in the velocity of the object.

In the given diagram, the resultant velocity of the puck is calculated as follows;

Figure a:

\Delta v = v_f -v_i\\\\\Delta v = 5 - 6 = - 1 \ m/s

Figure b:

v = \sqrt{4^2 + 3^2} \\\\v = 5 \ m/s

Figure c:

\Delta v = 4 - (-2)\\\\\Delta v = 6 \ m/s

Thus, the work done on the Puck by the applied force from the most positive to the most negative is c, b, a respectively.

Learn more here:brainly.com/question/19498865

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Answer:

Explanation:

Length = 1.00 m

If the length is 1.0, the vertical distance pivot to bob is cos 35 = 0.819

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12. A volcano erupts on Earth's surface causing a piece of rock with a mass of 95 kg to be ejected from rest to an acceleration
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Answer:

The force exerted on the rock by the eruption is, D. 902.5 N

Explanation:

Given data,

The mass of the rock ejected by the volcano, m = 95 kg

The acceleration of the ejected rock, a 9.5 m/s²

The force acting on an object is defined as the product of the mass and its acceleration. It is given by the relation,

                                 F = m x a

                                    = 95 x 9.5

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MArishka [77]

Answer:

1) a block going down a slope

2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c)  W = ΔK, d) ΔU = ΔK

Explanation:

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1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.

2)

a) rolling a ball uphill

In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact

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b) in this system work is transformed into internal energy

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