the focal length <span> is much more decent for a concave, and also worse</span><span> for a convex mirror. When the image that is given, distance is good and decent, images are always on the same area of the mirror as the object given , and it is not fake. images distance is </span>never positive <span>, the image is on the oppisite side of the mirror, so the image must be virtual.</span>
Answer:
both experience forces or at least a force
Explanation:
it would go in the direction the other object
(second object, the one that crashed) was going
si if going right then right if left then left
plus or minus
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>
Concept:
Frequency- It is defined as the number of oscillations occur in one second.
Its SI unit is Hertz (Hz)
Given: Produced sound vibrations is 18,500 cycles in 0.75 seconds
∵ In 0.75 second, produced sound has oscillations = 18,500 cycles
∴ In 1.0 second, produced sound has oscillations = (18,500 ÷ 0.75) Hz
The frequency of the sound will be ≈ 24,667 Hz
From the study of the given graph, only the animals (c) Cats, (b) Moths and (a) Bats can hear the produced sound because their upper audible frequency range is greater than 24,667 Hz.