A, an iconic ion covalently bonded with a metal
Answer:
Work done by the frictional force is
Explanation:
It is given that,
Mass of the car, m = 1000 kg
Initial velocity of car, u = 26.1 m/s
Finally, it comes to rest, v = 0
We have to find the work done by the frictional forces. Work done is equal to the change in kinetic energy as per work - energy theorem i.e.
W = −340605 J
or
Hence, the correct option is (a).
It can hurt and harm the body on many different and levels
Answer:
Frictional force, F = 45.9 N
Explanation:
It is given that,
Weight of the box, W = 150 N
Acceleration,
The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.
It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,
Frictional force is given by :
F = 45.9 N
So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.
I know i did part a correctly. heres what i did: momentum is conserved: m1 * u - m2 * u = m2 * v or (m1 - m2) * u = m2 * v Also, for an elastic head-on collision, we know that the relative velocity of approach = relative velocity of separation (from conservation of energy), or, for this problem, 2u = v Then (m1 - m2) * u = m2 * 2u m1 - m2 = 2 * m2 m1 = 3 * m2 m1 is the sphere that remained at rest (hence its absence from the RHS), so m2 = 0.3kg / 3 m2 = 0.1 kg b) this part confuses me, heres what i did (m1 - m2) * u = m2 * v (.3kg - .1kg)(2.0m/s) = .1kg * v .4 kg = .1 v v = 4 m/s What my teacher did: (.3g - .1g) * 2.0m/s = (.3g + .1g) * v I understand the left hand side but i dont get the right hand side. Why is m1 added to m2 when m1 is at rest which makes its v = zero?? v = +1.00m/s since the answer is positive, what does that mean? Also, if v was -1.00m/s what would that mean? thanks!
<span>Reference https://www.physicsforums.com/threads/elastic-collision-with-conservation-of-momentum-problem.651261...</span>