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:
none of the above
Explanation:
im pretty positive this is the answer tell me if i am wrong please
The only correct statement on that list is <em>choice-C</em>: If a positively charged rod is brought close to a positively charged object, the two objects will repel.
Answer: mechanical energy
Explanation: I hope this helps! Also, please mark as brainliest, thanks!
Answer:
a. 0.342 kg-m² b. 2.0728 kg-m²
Explanation:
a. Since the skater is assumed to be a cylinder, the moment of inertia of a cylinder is I = 1/2MR² where M = mass of cylinder and r = radius of cylinder. Now, here, M = 56.5 kg and r = 0.11 m
I = 1/2MR²
= 1/2 × 56.5 kg × (0.11 m)²
= 0.342 kgm²
So the moment of inertia of the skater is
b. Let the moment of inertia of each arm be I'. So the moment of inertia of each arm relative to the axis through the center of mass is (since they are long rods)
I' = 1/12ml² + mh² where m = mass of arm = 0.05M, l = length of arm = 0.875 m and h = distance of center of mass of the arm from the center of mass of the cylindrical body = R/2 + l/2 = (R + l)/2 = (0.11 m + 0.875 m)/2 = 0.985 m/2 = 0.4925 m
I' = 1/12 × 0.05 × 56.5 kg × (0.875 m)² + 0.05 × 56.5 kg × (0.4925 m)²
= 0.1802 kg-m² + 0.6852 kg-m²
= 0.8654 kg-m²
The total moment of inertia from both arms is thus I'' = 2I' = 1.7308 kg-m².
So, the moment of inertia of the skater with the arms extended is thus I₀ = I + I'' = 0.342 kg-m² + 1.7308 kg-m² = 2.0728 kg-m²
