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marysya [2.9K]
3 years ago
7

A vibrating tuning fork of frequency 512 Hz is held over a water column with one end closed and the other open. As the water lev

el is allowed to fall, a loud sound (resonance) is heard at specific water levels. Assume you start with the tube full of water, and begin steadily lowering the water level. What is the water level (as measured from the top of the tube) for the third such resonance?
Physics
1 answer:
aleksklad [387]3 years ago
8 0

Answer:

Explanation:

This is an example of resonance of air column with one end closed.

For third resonance , length of water column

= 5  x λ / 4

λ ( wave length ) = velocity / frequency

343 / 512

= .67 m

= 67 cm

length of air column

= 5 x λ / 4

= 5/4 x 67 cm

= 83.75 cm

water level will be 83.75 cm below.

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Name 2 common methods of polymerization.
navik [9.2K]

Answer: Addition polymerization & Condensation polymerization

6 0
3 years ago
Give THREE examples of objects that have potential energy
Tasya [4]

A ball kept on 3rd floor of a building.

A pendulum bob kept at 3m height

A stone thrown vertically upward.

A pressed spring.

A squashed spunge ball.

4 0
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Neptune is approximately 4.5 billion kilometers from the sun. What is this distance in AU?
asambeis [7]
Well, one AU is 149,597,870 km. So, we would basically have to divide 4.5 billion km by 149,597,870, right?

4,500,000,000/149,597,870=30.080642 AU.

So, the correct answer would be 30 AU. Hoped this helped!
7 0
3 years ago
An automobile rounds a curve of radius 50.0 m on a flat road.
bixtya [17]

Answer:

14m/s

Explanation:

Given parameters:

Radius of the curve  = 50m

Centripetal acceleration  = 3.92m/s²

Unknown:

Speed needed to keep the car on the curve = ?

Solution:

The centripetal acceleration is the inwardly directly acceleration needed to keep a body along a curved path.

 It is given as;

      a = \frac{v^{2} }{r}  

a is the centripetal acceleration

v is the speed

r is the radius

  Now insert the parameters and find v;

         v²   = ar

        v² = 3.92 x 50  = 196

         v  = √196 = 14m/s

6 0
3 years ago
A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal
Gemiola [76]

Answer:

Time taken, T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}

Explanation:

It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.

From the figure,

The sum of forces in y direction is :

T\ cos\theta-mg=0

T=\dfrac{mg}{cos\theta}

Sum of forces in x direction,

T\ sin\theta=\dfrac{mv^2}{r}

mg\ tan\theta=\dfrac{mv^2}{r}.............(1)

Also, r=l\ sin\theta

Equation (1) becomes :

mg\ tan\theta=\dfrac{mv^2}{l\ sin\theta}

v=\sqrt{gl\ tan\theta.sin\theta}...............(2)

Let t is the time taken for the ball to rotate once around the axis. It is given by :

T=\dfrac{2\pi r}{v}

Put the value of T from equation (2) to the above expression:

T=\dfrac{2\pi r}{\sqrt{gl\ tan\theta.sin\theta}}

T=\dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta.sin\theta}}

On solving above equation :

T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}

Hence, this is the required solution.

4 0
3 years ago
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