Answer: The free - body diagrams for blocks A and B. frictionless surface by a constant horizontal force F = 100 N. Find the tension in the cord between the 5 kg and 10 kg blocks. The string that attaches it to the block of mass M2 passes over a frictionless pulley of negligible mass. The coefficient of kinetic friction Hk between M.
Explanation: Hope this helped :)
If it takes 1 year for this crab to travel 5,70km
Then, it will takes approximatly 1000/5,70 = 175 years to travel 1000 km
By using the second law of Newton, the frictional force is 200N.
We need to know about the second law of Newton (force) to solve this problem. The total force applied an object is proportional to the mass of object and acceleration. It can be defined as
∑F = m . a
where F is force, m is mass and a is acceleration.
From the question above, we know that
F1 = 200N
v = constant therefore (a = 0 m/s²)
By using second law of Newton, we get
∑F = m . a
F1 - Ffriction = m . 0
200 - Ffriction = 0
Ffriction = 200 N
Hence, the frictional force is 200N.
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Answer:
i don't know if this is good for you but
Explanation:
ignoring frictional air resistance (drag) the speed on return is the same as when it left the ground (5 m/s but in the opposite direction).
Note: this points out a good reason for not firing live bullets into the air..they will return somewhere and at the same speed.
However, if you take into account the atmospheric drag the reurn speed will be somewhat smaller (but in the case of a bullet, probably still lethal.) Drag depends on many factors and is difficult to calculate.
the answer should be 20 seconds
