Answer:
a = 0.8 m/s^2
Explanation:
Force equation: F = ma
F = ma -> a = F/m = 2.8*10^3 N / 3.5*10^3 kg = 0.8 m/s^2
A. An electron has far less mass than either a proton or neutron.
<span>The person is dragging
with a force of 58 lbs at an angle of 27 degrees relating to the ground. We
want to use cosine function to look for the horizontal force component. And
then we can compute for W = (Horizontal Force) x (Distance). We want the
horizontal force component since that is the component that is parallel to the
direction the cart is moving. </span><span>
(cos 27 degrees)(58 lbs) = 51.69 lbs (This is the horizontal
force component.)
W = (51.69 lbs) x (70 ft) = 3618.3 ft*lbs</span>
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>
To do this we may use things that are good conductors - are painted dull black -
Have a air flow around them Maximised.
