Answer:
C
Explanation:
horizintal speed stays same
only vertical speed changes
so H speed will stay 30 m/s
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:
The required angular speed ω of an ultra-centrifuge is:
ω = 18074 rad/sec
Explanation:
Given that:
Radius = r = 1.8 cm
Acceleration due to g = a = 6.0 x 10⁵ g
Sol:
We know that
Angular Acceleration = Angular Radius x Speed²
a = r x ω ²
Putting the values
6 x 10⁵ g = 1.8 cm x ω ²
Converting 1.8 cm to 0.018 m, also g = 9.8 ms⁻²
6 x 10⁵ x 9.8 = 0.018 x ω ²
ω ² = (6 x 10⁵ x 9.8) / 0.018
ω ² = 5880000 / 0.018
ω ² = 326666667
ω = 18074 rad/sec
I think that it’s the first one
Answer:
Explanation:
The former is the case of constructive interference and the later case relates to destructive interference.
For listener, path difference is the separation of loudspeaker as listener is not standing in between the speaker.
If λ be the wave length
For constructive interference
19 = n λ
For destructive interference
29 = (2n+1) λ / 2
= n λ + λ / 2
= 19 + λ / 2
10 = λ / 2
λ = 20 cm
= 0. 20 m
