The speed of the second mass after it has moved ℎ=2.47 meters will be 1.09 m/s approximately
<h3>
What are we to consider in equilibrium ?</h3>
Whenever the friction in the pulley is negligible, the two blocks will accelerate at the same magnitude. Also, the tension at both sides will be the same.
Given that a large mass m1=5.75 kg and is attached to a smaller mass m2=3.53 kg by a string and the mass of the pulley and string are negligible compared to the other two masses. Mass 1 is started with an initial downward speed of 2.13 m/s.
The acceleration at which they will both move will be;
a = ( - ) / ( + )
a = (5.75 - 3.53) / (5.75 + 3.53)
a = 2.22 / 9.28
a = 0.24 m/s²
Let us assume that the second mass starts from rest, and the distance covered is the h = 2.47 m
We can use third equation of motion to calculate the speed of mass 2 after it has moved ℎ=2.47 meters.
v² = u² + 2as
since u =0
v² = 2 × 0.24 × 2.47
v² = 1.1856
v = √1.19
v = 1.0888 m/s
Therefore, the speed of mass 2 after it has moved ℎ=2.47 meters will be 1.09 m/s approximately
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D. have particle moment perpendicular to the direction of the energy.
The distance traveled by the hockey player is 0.025 m.
<h3>The principle of conservation of linear momentum;</h3>
- The principle of conservation of linear momentum states that, the total momentum of an isolated system is always conserved.
The final velocity of the hockey play is calculated by applying the principle of conservation of linear momentum;
The time taken for the puck to reach 15 m is calculated as follows;
The distance traveled by the hockey player at the calculated time is;
Learn more about conservation of linear momentum here: brainly.com/question/7538238
Explanation:
Sorry but I don't Understand question