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tankabanditka [31]
3 years ago
6

A. Label the following properties on the wave below: rarefaction, wavelength, and compression. (Image attached)

Physics
2 answers:
DedPeter [7]3 years ago
0 0

PART a)

Rarefraction : It is the region of lower density where particles move away from each other

Wavelength : distance between two consecutive compression or rarefaction is known as wavelength

Compression : The region of higher density or the region where particles comes closer to each other is compression

Part b)

Here it is given that

f = 2100 Hz

\lambda = 3.0 m

So here we know that

wave speed = frequency \times wavelength

[tex]v = 2100(3) = 6300 m/s

PART c)

As the person drop the stone in the water it creates waves on the surface of water

These waves can move the boat up and down at its own position and it can not displace the boat so it will not move to the shore

PART D)

When stone is dropped in the water it creates waves. these waves when travel through the water and disturb the boat

Due to this disturbance boat starts moving up and down at its own position and it will not displace from its position.

So by this type of waves it will not move towards the shore

Alecsey [184]3 years ago
0 0

a) Rarefactions are in the green frames, compressions in the purple. Wavelength is a distance between same points of the two corresponding rarefactions/compressions.

b) v=wavelength times frequency=3*2100= 6300 m/s.

c) Bad idea that most likely won't work. Waves on water move only in vertical direction. But maybe a high wave will lift the boat and rescue it.

d) Ah, he needs wants the boat reach the shore, I thought it stuck and the person just wants to make it float. Yes, that's not the best idea, see explanation above

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There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.  

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The equilibrium condition for the mass 1 is

\displaystyle F_a-F_{r1}-F_{r2}=m.a=0

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The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

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The friction forces are computed by

\displaystyle F_{r2}=\mu \ N_2=\mu\ m_2\ g

\displaystyle F_{r1}=\mu \ N_1=\mu(m_1+m_2)g

Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.

Replacing in [1]

\displaystyle F_{a}=\mu \ m_2\ g\ +\mu(m_1+m_2)g

Simplifying

\displaystyle F_{a}=\mu \ g(m_1+2\ m_2)

Plugging in the values

\displaystyle F_{a}=0.25(9.8)[400+2(100)]

\boxed{F_a=1470\ N}

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