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3 years ago

6

When working with a powered instrument, the swirling effect produced by the water stream within the confined space of a periodon

tal pocket is termed _____.

1 answer:

Lina20 [59]3 years ago

3 0

Answer:

Acoustic microstreaming

Explanation:

Acoustic microstreaming is the swirling effect produced by water stream confined in a spaced of a periodontal pocket.

It is the movement of water in a particular direction as a result of mechanical pressure within the fluid body.
They are often used in dental procedures to remove particulates from the teeth.
It mostly relies on the properties of sound waves to achieve this goal

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Two blocks are connected by a light string that passes over two frictionless pulleys. The block of mass m2 is attached to a spri

irina1246 [14]

(BELOW YOU CAN FIND ATTACHED THE IMAGE OF THE SITUATION)

Answer:

$d=\frac{2g(m1-m2)}{k}$

Explanation:

For this we're going to use conservation of mechanical energy because there are nor dissipative forces as friction. So, the change on mechanical energy (E) should be zero, that means:

$E_{i}=E_{f}$

$K_{i}+U_{i}=K_{f}+U_{f}$ (1)

With $K_{i}$ the initial kinetic energy, $U_{i}$ the initial potential energy, $K_{f}$ the final kinetic energy and $U_{f}$ the final potential energy. Note that initialy the masses are at rest so $K_{i} = 0$ , when they are released the block 2 moves downward because m2>m1 and finally when the mass 2 reaches its maximum displacement the blocks will be instantly at rest so $K_{f} =0$ . So, equation (1) becomes:

$U_{i}=U_{f}$ (2)

At initial moment all the potential energy is gravitational because the spring is not stretched so $U_{i}=U_{gi}$ and at final moment we have potential gravitational energy and potential elastic energy so $U_{f}=U_{gf}+U_{ef}$ , using this on (2)

$U_{gi}=U_{gf}+U_{ef}$ (3)

Additional if we define the cero of potential gravitational energy as sketched on the figure below (See image attached), $U_{gi}=0$ and we have by (3) :

$0= U_{gf}+U_{ef}$ (4)

Now when the block 1 moves a distance d upward the block 2 moves downward a distance d too (to maintain a constant length of the rope) and the spring stretches a distance d, so (4) is:

$0=-m1gd+m2gd+\frac{kd^{2}}{2}$

dividing both sides by d

$0=-m1g+m2g+\frac{kd}{2}$

$g(m1-m2)= \frac{kd}{2}$

$d=\frac{2g(m1-m2)}{k}$ , with k the constant of the spring and g the gravitational acceleration.

7 0

3 years ago

A chimpanzee sitting against his favorite tree gets up and walks 70.9 m due east and 31.9 m due south to reach a termite mound,

DIA [1.3K]

A right triangle is formed by the 70.9 m walked east and 31.9 m walked south.
The legs of this right triangle are 70.9 and 31.9.
The shortest distance between two points is a straaight line. Therefore the hypotenuse of this triangle is going to be that shortest distance.
We can use the Pythagorean Theorem to find the hypotenuse of this triangle.
$a^2+b^2=c^2\\70.9^2+31.9^2=c^2\\5026.81+1017.61=c^2\\6044.42=c^2\\c=\sqrt{6044.42}$

$c=\sqrt{6044.42}\approx\boxed{77.7458681}\ (decimal\ form)\\\\c=\sqrt{6044.42}=\sqrt{\frac{604442}{100}}=\boxed{\frac{\sqrt{604442}}{10}}\ (exact\ form)$

As for the second part of the question, we want to find the angle formed by the hypotenuse and the 31.9 walked east.
We could use any of the three trigonometric ratios here since we know all 3 sides.
sine = opposite / hypotenuse
cosine = adjacent / hypotenuse
tangent = opposite / adjacent
I am going to use tangent, because then I won't have to deal with the hypotenuse and so the answer will be more accurate.

If you haven't already drawn yourself a diagram, now is a good time to.
The side opposite our angle is the 31.9, and the adjacent is 70.9.
Therefore, $\tan(m\angle)=\frac{31.9}{70.9}$ .

We can use inverse trig ratios here to find the measure of our angle.
$\tan^{-1}(\tan(m\angle))=\tan^{-1}(\frac{31.9}{70.9})\\\\m\angle=\tan^{-1}(\frac{31.9}{70.9})\approx\boxed{24.2243851\°\ or\ 0.422795279\ rad}$

4 0

3 years ago

How did planck find the correct curve for the specturm of light emitted by a hot obkect?

Neko [114]

Planck find the correct curve for the specturm of light emitted by a hot object by vibrational energies of the atomic resonators were quantized.

<h3>Briefing :</h3>

The energy density of a black body between λ and λ + dλ is the energy E=hc/λ of a mode times the density of states for photons, times the probability that the mode is occupied.
This is Planck's renowned equation for a black body's energy density.
According to this, electromagnetic radiation from heated bodies emits in discrete energy units or quanta, the size of which depends on a fundamental physical constant (Planck's constant). The basis of infrared imaging is the correlation between spectral emissivity, temperature, and radiant energy, which is made possible by Planck's equation.

Learn more about the Planck's constant with the help of the given link:

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#SPJ4

3 0

1 year ago

A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.3 m/s rel

Sonja [21]

consider east-west direction along X-axis and north-south direction along Y-axis

$V_{ra}$ = velocity of migrating robin relative to air = 12 j m/s

(where "j" is unit vector in Y-direction)

$V_{ag}$ = velocity of air relative to ground = 6.3 i m/s

(where "i" is unit vector in X-direction)

$V_{rg}$ = velocity of migrating robin relative to ground = ?

using the equation

$V_{rg}$ = $V_{ra}$ + $V_{ag}$

$V_{rg}$ = 12 j + 6.3 i

$V_{rg}$ = 6.3 i + 12 j

magnitude : sqrt((6.3)² + (12)²) = 13.6 m/s

direction : tan⁻¹(12/6.3) = 62.3 deg north of east

4 0

3 years ago

A launched hopper reach to 1.20 m maximum height. How much is it’s launch velocity?

garri49 [273]

The launch velocity is 4.8 m/s

Explanation:

We can solve this problem by applying the law of conservation of energy. In fact, the mechanical energy of the hopper (equal to the sum of the potential energy + the kinetic energy) is conserved. So we can write:

$U_i +K_i = U_f + K_f$

where:

$U_i$ is the initial potential energy, at the bottom

$K_i$ is the initial kinetic energy, at the bottom

$U_f$ is the final potential energy, at the top

$K_f$ is the final kinetic energy, at the top

We can rewrite the equation as:

$mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2$

where:

m is the mass of the hopper

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

$h_i = 0$ is the initial height

u is the launch speed of the hopper

$h_f = 1.20 m$ is the maximum altitude reached by the hopper

v = 0 is the final speed (which is zero when the hopper reaches the maximum height)

Solving the equation for u, we find the launch speed of the hopper:

$u=\sqrt{2gh_g}=\sqrt{2(9.8)(1.20)}=4.8 m/s$

Learn more about kinetic energy and potential energy:

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brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

4 0

3 years ago