5m
Explanation:
Given parameters:
Weight of object = 50N
Work done in lifting object = 250J
Unknown:
Vertical height = ?
Solution:
The work done on an object is the force applied to lift a body in a specific direction.
Work done = force x distance
Weight is a force in the presence of gravity;
Work done = weight x height of lifting
Height of lifting = 
Height of lifting = 
= 5m
The vertical height through which the object was lifted is 5m
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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>
Answer:hat are some examples of energy transformation?
The Sun transforms nuclear energy into heat and light energy.
Our bodies convert chemical energy in our food into mechanical energy for us to move.
An electric fan transforms electrical energy into kinetic energy.
Explanation:
Answer:
Work done by a tug boat, W = 1.735 x 10⁸ J
Explanation:
Given,
The of each tugboat, F = 1.5 x 10⁶ N
The angle of each tugboat forms with the resultant force, θ = 19°
The displacement of the supertanker, s = 710 m
The individual tugboat will be responsible for the displacement, d = 710/2
= 355 m
The displacement component in each tugboat direction = 355 · sin θ meter
Therefore, the work done by each tugboat is
W = F x S joules
Substituting the values in the above equation
W = 1.5 x 10⁶ x 355 · sin θ
= 1.735 x 10⁸ J
Hence, the work done by each tugboat is, W = 1.735 x 10⁸ J
