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wolverine [178]
1 year ago
11

A tube with a cap on one end, but open at the other end, produces a standing wave whose fundamental frequency is 130.8 Hz. The s

peed of sound is 343 m/s. a. If the cap is removed , what is the new fundamental frequency?
Physics
1 answer:
Angelina_Jolie [31]1 year ago
8 0

Answer:

A. 261.6 hz.

B. 0.656 m.

Explanation:

A.

When yhe tube is open at one end and closed at the other,

F1 = V/4*L

Where,

F1 = fundamental frequency

V = velocity

L = length of the tube

When the tube is open at both ends,

F'1 = V/2*L

Where

F'1 = the new fundamental frequency

Therefore,

V/2*L x V/4*L

F'1 = 2 * F1

= 2 * 130.8

= 261.6 hz.

B.

F1 = V/4*L

Or

F'1 = V/2*L

Given:

V = 343 m/s

F1 = 130.8

L = 343/(4 * 130.8)

= 0.656 m.

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In a distant solar system, a giant planet has
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Answer:

mass of the planet: 5.9\,10^{26}\,kg

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m\frac{v^2}{R}=G\frac{M\,m}{R^2}

where we used here the tangential velocity (v) of the moon around the planet. This equation can be further simplified by dividing both sides by "m" and multiplying both sides by the orbital radius R:

m\frac{v^2}{R}=G\frac{M\,m}{R^2}\\v^2=G\frac{M}{R}

Notice that the mass of the moon has actually disappeared from the equation, which tells us that the orbiting velocity and period do not depend on the mass of the moon, but on the mass of the actual planet.

We know the orbital radius R (5.32\,10^5\,km=5.32\,10^8\,m, the value of the Universal Gravitational constant G, and we can estimate the value of the tangential velocity of the moon since we know it period: 36.3 hrs = 388800 seconds.

We know that the moon makes a full circumference (2\,\pi\,R) in 388800 seconds, therefore its tangential velocity is:

v=\frac{2\,\pi\,5.32\,10^8}{388800} \frac{m}{s} \\v=8.6\,10^3\,\frac{m}{s}

where we rounded the velocity to one decimal.

Notice that we have converted all units to the SI system, so when using the formula to solve for the mass of the planet, the answer comes directly in kg.

Now we use this value for the tangential velocity to estimate the mass of the planet in the first equation we made and simplified:

v^2=G\frac{M}{R}\\M=\frac{v^2\,R}{G} \\M=\frac{(8.6\,10^3)^2\,5.32\,10^8}{6.67\,10^{-11}}kg\\M=5.9\,10^{26}\,kg

8 0
2 years ago
Some of the fastest dragsters (called "top fuel) do not race for more than 300-400m for safety reasons. Consider such a dragster
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Answer:

1.10261 times g

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Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow 400=0\times 8.6+\frac{1}{2}\times a\times 8.6^2\\\Rightarrow a=\frac{400\times 2}{8.6^2}\\\Rightarrow a=10.81665\ m/s^2

Dividing by g

\dfrac{a}{g}=\dfrac{10.81665}{9.81}\\\Rightarrow \dfrac{a}{g}=1.10261\\\Rightarrow a=1.10261g

The acceleration is 1.10261 times g

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186.04644\times \dfrac{3600}{1609.34}=416.17506\ mph

The speed of the dragster is 416.17506 mph

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