Answer:
<em>Good Luck!</em>
Explanation:
Force applied by first student (F1) = 80 N
Force applied by the second student (F2) = 40 N
Displacement (d) = 10 m
Work done by first student (W1) = F1d = 80*10 = 800 J
Work done by second student (W2) = F2d = 40*10 = 400 J
Hence, the work done by the pushing student is 800 J and that by pulling student is
<h2> <u><em>400 J</em></u></h2>
Answer:
15.6m/s
Completed Question;
For a short period of time, the frictional driving force acting on the wheels of the 2.5-Mg van is N= 600t^2 , where t is in seconds. If the van has a speed of 20 km/h when t = 0, determine its speed when t = 5
Explanation:
Mass m = 2500kg
Speed v1 = 20km/h = 20/3.6 m/s = 5.556 m/s
To determine speed v2;
Using the principle of momentum and impulse;
mv1 + ∫₀⁵ F dt = mv2
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<u>Answer
</u>
4.The objects have the same acceleration
<u>Explanation
</u>
Acceleration due to gravity is the acceleration acquired by an object when falling freely.
Under normal conditions, different objects has different accelerations depending on their mass and surface area. This is brought about by the air resistance they experience as they fall.
The case is difference when they the experiment is carried out in a controlled condition. That is in vacuum. I this case the acceleration of the objects is the same.
