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shutvik [7]
4 years ago
8

If the above two waveforms were sound waves, we would hear the ___________ wave louder. If the above two waveforms were light wa

ves, we would see the ________ wave dimmer.
Physics
1 answer:
murzikaleks [220]4 years ago
7 0
To compute the net effect of two waves, we use the superposition principle, and we can call the resultant wave "superposed wave".
We can rewrite the sentence as follows:
"<span>If the above two waveforms were sound waves, we would hear the superposed wave louder. If the above two waveforms were light waves, we would see the superposed wave dimmer."</span>
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A system absorbs 194 kj of heat and the surroundings do 120 kj of work on the system. internal eneergy change
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We can solve the problem by using the first law of thermodynamics, which states that:
\Delta U = Q-W
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\Delta U is the change in internal energy of the system
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Why is the mechanical advantage of using a single pulley always 1? Assume there’s no friction. A. The input force is in a differ
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4 years ago
Read 2 more answers
(d) Suppose you use a spring to launch a payload horizontally from the asteroid so that the payload ends up far from the asteroi
nydimaria [60]

Answer:

ks= 133.2 N/m

Explanation:

  • Assuming that we can neglect the gravitational potential energy of the mass, and that no other forces acting on the payload, total mechanical energy must be conserved.
  • This energy, at any time, is part elastic potential energy (stored in the spring) and part kinetic energy.
  • When the spring is initially compressed, the payload is at rest, so all energy is elastic potential.
  • Once the spring has returned to its natural state, all this elastic potential energy must have been turned into kinetic energy.
  • If the payload is launched horizontally, and no gravity is present,this means that its final speed will be horizontal only also, according to Newton's First Law.
  • So, we can write the following equation:

       \Delta U + \Delta K = 0 (1)

  • where ΔU = -1/2*k*(Δx)²  (2)
  • and ΔK = 1/2*m*v² (3)
  • Replacing in (2) and (3) by the givens, and simplifying, we can find the stiffness ks as follows:

       k_{s} =\frac{m*v^{2}}{\Delta x^{2}} = \frac{29 kg*(3m/s)^{2}}{(1.4m)^{2}} = 133.2 N/M (4)

5 0
3 years ago
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