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Maksim231197 [3]
3 years ago
8

An organ pipe of length 3.0 m has one end closed. The longest and next-longest possible wavelengths for standing waves inside th

e pipe are
Physics
1 answer:
iris [78.8K]3 years ago
6 0

Answer:

The longest wavelength for closed at one end and open at the other is

y / 4      where y is the wavelength - that is node - antinode

The next possible wavelength is 3 y / 4 -    node - antinode - node -antinode

y / 4 = 3 m     y = 12 meters    the longest wavelength

3 y / 4 = 3 m      y = 4 meters   1 / 3 times as long

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F an object has a mass of 200 kg and a weight of 1000 N, what is g?
jeka94

Answer:

g = 5 m/s square

Explanation:

Weight(W), Mass(m), Gravity(g)

W = mg

1,000N = 200g

g = 1000/200

g = 5 m/s square

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3 years ago
4. I drop a pufferfish of mass 5 kg from a height of 5.5 m onto an upright spring of total length 0.5 m and spring constant 3000
KatRina [158]

Answer:

a)  0.28 m or 28 cm is the minimum  height above ground the fish reaches.

b)  at the height of 0.484 m height , the pufferfish will eventually come to rest.

c) There exists  two types of energy remain at the equilibrium point in the system. These are :

Gravitational potential energy  = 23.72J

Spring potential energy   = 0.384 J

Explanation:

Given that :

Mass of the pufferfish m =5kg

initial height of the fish h =5.5m

length of the spring l =0.5m

Spring constant K =3000N/m

a)

Assuming no energy loss to friction, what is the minimum height above the ground that the pufferfish reaches?

Lets assume that the minimum height the fish reaches is = x meters

Now by using the conservation of energy; we realize that :

Initial total energy = final total energy

Gravitational potential energy =

Gravitational potential energy' + Spring potential energy (kinetic energy is zero in both cases)

mgh = mgx + \frac{1}{2}K(l-x)^2

Replacing our given values into the above equation; we have :

(5)(9.8)(5.5) = (5)(9.5)(x) + \frac{1}{2}(3000)(0.5-x)^2

269.5 = 47.5 x + 1500(0.5 -x )²

269.5 = 47.5 x + 1500(0.25 - x²)

269.5 = 47.5 x + 375 - 1500 x²

269.5 - 375 = 47.5 x - 1500 x²

-105.5 = 47.5 x - 1500 x²

-105.5 + 1500 x² - 47.5 x = 0

1500 x² - 47.5 x - 105.5 = 0

By using quadratic equation and taking the positive value;

x = 0.28 m or 28 cm is the minimum height above ground the fish reaches.

b)

At the equilibrium position the weight of fish will be equal to the force applied by the spring thus

mg = kx

substituting  our given values ; we have:

(5)(9.8) = 3000x

x = 61.22

x = 0.016m  : so this is the compression in the spring

Now; to determine the height  the pufferfish gets to before  it eventually come to rest; we have

(0.5-0.016) m = 0.484m

therefore, at the height of 0.484 m height , the pufferfish will eventually come to rest.

c)

There exists  two types of energy remain at the equilibrium point in the system. These are :

Gravitational potential energy  = mgh' = (5)(9.8)(0.484)

= 23.72J

and spring potential energy  

=\frac{1}{2}Kx^2\\ = \frac{1}{2}(3000)(0.016)^2\\= 0.384J

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simple

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