Answer:
If the line is curved, the slope is changing, which also means the velocity is changing. In a distance-time graph, the gradient of the line is equal to the speed of the object. The more the gradient (and the steeper the line) the faster the object is moving.
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:
A. Voltage
Explanation:
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Answer:
The answer to your question is: a = 1.99 m/s²
Explanation:
Data
mass = 20 kg
angle = 56°
Force = 71 N
horizontal acceleration = ?
Process
Find the horizontal force
cos Ф = adjacent side / hypotenuse
adjacent side = hypotenuse x cosФ
adjacent side = 71 x cos 56
a.s. = 39.70 N
Newton's second law
F = ma
a = F/m
a = 39.7 / 20
a = 1.99 m/s²