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Lady bird [3.3K]
3 years ago
15

Which choice is a valid way to construct a motion diagram?

Physics
1 answer:
tigry1 [53]3 years ago
8 0

Answer:

A valid way to construct a motion diagram is to take a series of photographs of a moving object at equal time intervals of time, perpendicular to the direction of motion and overlaying the images to observe the position change with time.

Explanation:

In order to develop a motion diagram, a coordinate system, a zero and appropriate scale should be chosen. Generally positions to the right of the origin are considered positive. Velocity can be determined from the motion diagram by plotting velocity vectors.

The acceleration of an object can also be calculated from the motion diagram.

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A cat with mass 4.0 kg jumps down to the floor from a bookcase 2.0 m high. What is the cat's change in
eduard

Answer: -7J

Explanation:

To determine the change in potential energy, use the equation ΔPE=mgΔh

Δ

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, where m is the mass, g= 9.8 m/s2

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, and Δh is the change in height. Hence, we have that the potential energy PE=4.0 kg×9.8 m/s2×-2.0 m=-78 J

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. To verify that this is correct, note that since the cat changes the potential energy to kinetic energy by jumping, the potential energy decreases. Hence, the potential energy should be negative.

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Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 25 kg and the larger bottom crate has a m
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Answer:

Explanation:

Let the bigger crate be in touch with the ground which is friction less. In the first case both m₁ and m₂ will move with common acceleration because m₁ is not sliding over m₂.

1 ) Common acceleration a = force / total mass

= 234 / ( 25 +91 )

= 2.017 m s⁻².

2 ) Force on m₁ accelerating it , which is nothing but friction force on it by m₂

= mass x acceleration

= 25 x 2.017

= 50.425 N

The same force will be applied by m₁ on m₂ as friction force which will act in opposite direction.

3 ) Maximum friction force that is possible between m₁ and m₂

= μ_s m₁g

= .79 x 25 x 9.8

= 193.55 N

Acceleration of m₁

= 193 .55 / 25

= 7.742 m s⁻²

This is the common acceleration in case of maximum tension required

So tension in rope

= ( 25 +91 ) x 7.742

= 898 N

4 ) In case of upper crate sliding on m₂ , maximum friction force on m₁

=  μ_k m₁g

= .62 x 25 x 9.8

= 151.9 N

Acceleration of m₁

= 151.9 / 25

= 6.076 m s⁻².

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3 years ago
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Answer:

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