Ask question

Ask question

All categories

3 years ago

9

Laura and Elana are discussing how to solve the following problem: "A canary sits 10 m from the end of a 30-m-long clothesline,

and a grackle sits 5 m from the other end. The rope is pulled by two poles that each exert a 200-N force on it. The mass per unit length is 0.10 kg/m. At what frequency must you vibrate the line in order to dislodge the grackle while allowing the canary to sit undisturbed?"

1 answer:

horrorfan [7]3 years ago

3 0

Answer:

$f=2.236\ Hz$

Explanation:

Given:

Length of a rope, $l=30\ m$

Position of Canary on the rope from one end, $l_c=10\ m$

Position of Grackle on the rope from another end, $l_g=5\ m$

Tension in the rope, $F_T=200\ N$

linear mass distribution on the rope, $\mu=0.1\ kg.m^{-1}$

We have for the speed of wave on the string:

$v^2=\frac{F_T}{\mu}$

$v^2=\frac{200}{\0.1}$

$v=44.7\ m.s^{-1}$

<em>For canary to be undisturbed we need a node at this location.</em>

<em>Also, at the end close to Canary there must be a node to avoid any change in pattern of vibration.</em>

So,

the distance between Canary and the closer end must be equal to half the wavelength.

$\frac{\lambda}{2} =10\ m$

$\Rightarrow \lambda=20\ m$

∴Wavelength of wave to be produced = 20 m. This will give us nodes at the multiples of 10 and anti-nodes at the multiples of 5.

Now, frequency:

$f=\frac{v}{\lambda}$

$f=\frac{44.7}{20}$

$f=2.236\ Hz$

You might be interested in

HELPPPPP WILL GIVE BRAINLIEST!!!

Lady bird [3.3K]

Answer:

3.round object that orbits the Sun but lacks the ability to clear the neighborhood around its orbit.

Explanation:

in 2006 the IAU, said that a dwarf planet is round object that has not cleared the area round a object and that is why Pluto, Ceres, and Eris are dwarf planet.

7 0

3 years ago

Read 2 more answers

If an area has all the wolves that it can support the wolf population has reached is what

Liono4ka [1.6K]

The wolf population in that area has reached its carrying capacity.

7 0

3 years ago

Read 2 more answers

Read the passage.

Paladinen [302]

<h2>Answer: B)Scientists’ understanding of cells continually improved as the results of studies built upon each other over time and formed the cell theory.</h2>

Explanation:

Nowadays we know <u>cells are essential microscopic units that make up the living beings, capable of reproducing independently. </u>

However, this is the result of a long process of discoveries and studies made since the 19th century, in which the continuous improvement of new technologies was helpful.

In fact, it is wel known the English scientist Robert Hooke was the first to discover the existence of cells by looking through a compound microscope at a cork sheet, realizing that it was made up of small polygonal holes (like those of a honeycomb) that reminded him of the chambers in which the monks stayed (called cells). Then, during the next centuries more studies were made until we had the current knowledge about the structure of a cell.

7 0

3 years ago

50 points !! I need help asap.......Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to t

r-ruslan [8.4K]

1) At the top of the building, the ball has more potential energy

2) When the ball is halfway through the fall, the potential energy and the kinetic energy are equal

3) Before hitting the ground, the ball has more kinetic energy

4) The potential energy at the top of the building is 784 J

5) The potential energy halfway through the fall is 392 J

6) The kinetic energy halfway through the fall is 392 J

7) The kinetic energy just before hitting the ground is 784 J

Explanation:

1)

The potential energy of an object is given by

$PE=mgh$

where

m is the mass

g is the acceleration of gravity

h is the height relative to the ground

While the kinetic energy is given by

$KE=\frac{1}{2}mv^2$

where v is the speed of the object

When the ball is sitting on the top of the building, we have

$h=40 m$ , therefore the potential energy is not zero
$v=0$ , since the ball is at rest, therefore the kinetic energy is zero

This means that the ball has more potential energy than kinetic energy.

2)

When the ball is halfway through the fall, the height is

$h=20 m$

So, half of its initial height. This also means that the potential energy is now half of the potential energy at the top (because potential energy is directly proportional to the height).

The total mechanical energy of the ball, which is conserved, is the sum of potential and kinetic energy:

$E=PE+KE=const.$

At the top of the building,

$E=PE_{top}$

While halfway through the fall,

$PE_{half}=\frac{PE_{top}}{2}=\frac{E}{2}$

And the mechanical energy is

$E=PE_{half} + KE_{half} = \frac{PE_{top}}{2}+KE_{half}=\frac{E}{2}+KE_{half}$

which means

$KE_{half}=\frac{E}{2}$

So, when the ball is halfway through the fall, the potential energy and the kinetic energy are equal, and they are both half of the total energy.

3)

Just before the ball hits the ground, the situation is the following:

The height of the ball relative to the ground is now zero: $h=0$ . This means that the potential energy of the ball is zero: $PE=0$

The kinetic energy, instead, is not zero: in fact, the ball has gained speed during the fall, so $v\neq 0$ , and therefore the kinetic energy is not zero

Therefore, just before the ball hits the ground, it has more kinetic energy than potential energy.

4)

The potential energy of the ball as it sits on top of the building is given by

$PE=mgh$

where:

m = 2 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity

h = 40 m is the height of the building, where the ball is located

Substituting the values, we find the potential energy of the ball at the top of the building:

$PE=(2)(9.8)(40)=784 J$

5)

The potential energy of the ball as it is halfway through the fall is given by

$PE=mgh$

where:

m = 2 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity

h = 20 m is the height of the ball relative to the ground

Substituting the values, we find the potential energy of the ball halfway through the fall:

$PE=(2)(9.8)(20)=392 J$

6)

The kinetic energy of the ball halfway through the fall is given by

$KE=\frac{1}{2}mv^2$

where

m = 2 kg is the mass of the ball

v = 19.8 m/s is the speed of the ball when it is halfway through the fall

Substituting the values into the equation, we find the kinetic energy of the ball when it is halfway through the fall:

$KE=\frac{1}{2}(2)(19.8)^2=392 J$

We notice that halfway through the fall, half of the initial potential energy has converted into kinetic energy.

7)

The kinetic energy of the ball just before hitting the ground is given by

$KE=\frac{1}{2}mv^2$

where:

m = 2 kg is the mass of the ball

v = 28 m/s is the speed of the ball just before hitting the ground

Substituting the values into the equation, we find the kinetic energy of the ball just before hitting the ground:

$KE=\frac{1}{2}(2)(28)^2=784 J$

We notice that when the ball is about to hit the ground, all the potential energy has converted into kinetic energy.

Learn more about kinetic and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

4 0

3 years ago

Does the mass of a pendulum affects the period of oscillation

pshichka [43]

Mass does not affect the pendulum's swing. The longer the length of string, the farther the pendulum falls; and therefore, the longer the period, or back and forth swing of the pendulum. The greater the amplitude, or angle, the farther the pendulum falls; and therefore, the longer the period.

8 0

3 years ago