Consider a system to be one train car moving toward another train car at rest. When the train cars collide, the two cars stick t
ogether.
What is the total momentum of the system after the collision?
800 kg • m/s
1,600 kg • m/s
2,400 kg • m/s
4,000 kg • m/s
What is the total momentum of the system after the collision?
800 kg • m/s
1,600 kg • m/s
2,400 kg • m/s
4,000 kg • m/s
2 answers:
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Answer:
<u>The pendulum bob swing past the mean position because:</u>
When a pendulum's bob is accelerating at its extreme position its velocity is zero. Due to the restoring toque the bob starts to accelerates towards its mean postion. The maximum acceleration of the pendulum's bob is and the the acceleration decreases as
towards the mean position.
The acceleration at the mean position becomes zero but the velocity remains maximum. Hence the bob continues to move and does not stops.Thus it can summarised as the force decreases ,acceleration decreases and velocity increases at slow rate.
Answer: The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increased by 2.0 kg, the period is found to be 3.0 s. Then the mass m is 0.625kg.
Explanation: To find the answer, we need to know more about the simple harmonic motion.
<h3>What is simple harmonic motion?</h3>- A particle is said to execute SHM, if it moves to and fro about the mean position under the action of restoring force.
- We have the equation of time period of a SHM as,
- Where, m is the mass of the body and k is the spring constant.
- Given that,
- We have to find the value of m,
Thus, we can conclude that, the mass m will be 0.625kg.
Learn more about simple harmonic motion here:
#SPJ4
To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.
In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.
By definition we know that the tensile strength is defined as
Where,
Tensile strength
F = Tensile Force
A = Cross-sectional Area
In the other hand we have that the shear strength is defined as
where,
Shear strength
Shear Force
Parallel Area
PART A) Replacing with our values in the equation of tensile strenght, then
Resolving for F,
PART B) We need here to apply the shear strength equation, then
In such a way that the material is more resistant to tensile strength than shear force.