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3 years ago

A 2.0-kg ball has a momentum of 25kg.m/s what is the ball's speed?

Physics

2 answers:

Reil [10]3 years ago

5 0

We know, momentum = mass * speed

25kgm/s = 2 kg * s

s = 25/2 = 12.5 m/s

Mazyrski [523]3 years ago

5 0

The ball's speed is about 13 m/s

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<h3>Further explanation</h3>

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

$\large {\boxed {F = ma }$

<em>F = Force ( Newton )</em>

<em>m = Object's Mass ( kg )</em>

<em>a = Acceleration ( m )</em>

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$\large {\boxed {F = \frac{\Delta p}{t} }$

<em>F = Force ( Newton )</em>

<em>p = Momentum of Object ( kg m/s )</em>

<em>t = Time Taken ( s )</em>

Let us now tackle the problem !

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<u>Given:</u>

mass of ball = m = 2.0 kg

momentum of ball = p = 25 kgm/s

<u>Asked:</u>

speed of the ball = ?

<u>Solution:</u>

<em>We could use formula for </em><em>momentum</em><em> to solve this problem:</em>

$p = m v$

$v = p \div m$

$v = 25 \div 2.0$

$v = 12.5 \texttt{ m/s}$

$v \approx 13 \texttt{ m/s}$

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<h3>Conclusion :</h3>

The ball's speed is about 13 m/s

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<h3>Learn more</h3>

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441
Newton's Law of Motion: brainly.com/question/10431582
Example of Newton's Law: brainly.com/question/498822

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

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Answer:

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A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The

USPshnik [31]

Answer:

a) see attached, a = g sin θ

c) v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

Wₓ = m a

W sin θ = m a

a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

θ' = 9/2 θ

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

Em₀ = mg h = mg L (1-cos tea)

Lowest point

Emf = K = ½ m v²

Em₀ = Emf

g L (1-cos θ) = v² / 2

v = √(2gL (1-cos θ))