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3 years ago

Discuss the role of the expanded functions dental assistant in making a provisional coverage

1 answer:

7 0

The expanded-function dental assistant (EFDA) can play a major role in the fabrication and temporary cementation of a provisional crown or bridge. It is the dentist’s and the EFDA’s responsibility to remain current with the new provisional materials and techniques that are available. It is essential that a provisional crown or bridge remain cemented while the fixed prosthesis is being prepared and delivered to the dental office. When the patient returns for final cementation of a fixed crown or bridge, the provisional should be cautiously removed without causing any fracture or harm, just in case it will need to be recemented if the final prostheses needs to be sent back to the lab for adjustments and remake.

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A moving rope (parallel to the slope) is used to pull skiers up the mountain. If the slope of the hill is 37" and friction is ne

Charra [1.4K]

Since rope is parallel to the inclined plane so here we can say that net force parallel to the person which is pulling upwards must counterbalance the component of weight of the person.

Now here we will do the components of the weight of the person

given that weight of the person = 500 N

now its components are

$W_x = 500 cos37$

$W_y = 500 sin37$

now here as we can say that one of the component is balanced here by the normal force perpendicular to plane

while the other component of the weight is balanced by the force applied on the rope

So here the force applied on the rope will be given as

$F = W_y = 500 sin37$

$F = 300 N$

so it apply 300 N force along the inclined plane

5 0

3 years ago

Which of the following is the SI unit for volume

Katarina [22]

Meters for mass kilograms for volume cubic meters for density kilograms per cubic meter

3 0

2 years ago

Which of the following nuclei is most stable based on its binding energy?

Anettt [7]

We have that the most stable nuclei are the ones with the highest average binding energy. We see that Nitrogen has a mass number of 15 and that in this region of the graph average binding energy is low. Silver and Gold are along a line where there is a constant decline in average binding energy; silver has more than gold. However, we see that at the start of this decline, there is Fe 56. This region has the elements with the highest average binding energy; Nickel with a mass number of 58 is right there and thus it is the most stable nucleus out of the listed ones.

4 0

3 years ago

Consider an insulating sphere of radius 6 cm surrounded by a conducting sphere of inner radius 18 cm and outer radius 26 cm. Fur

dimulka [17.4K]

Answer:

$-1.7908787542\times 10^{-9}\ C$

$3.4260289211\times 10^{-9}\ C$

$1.7908787542\times 10^{-9}\ C$

$1.6351501669\times 10^{-9}\ C$

Explanation:

r = Radius

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Electric field is given by

$E=-\dfrac{kq}{r^2}\\\Rightarrow q=-\dfrac{Er^2}{k}\\\Rightarrow q=-\dfrac{1610\times 0.1^2}{8.99\times 10^9}\\\Rightarrow q=-1.7908787542\times 10^{-9}\ C$

The charge is $-1.7908787542\times 10^{-9}\ C$

$Q+q=\dfrac{Er^2}{k}\\\Rightarrow Q=\dfrac{Er^2}{k}-q\\\Rightarrow Q=\dfrac{120\times 0.35^2}{8.99\times 10^9}-(-1.7908787542\times 10^{-9})\\\Rightarrow Q=3.4260289211\times 10^{-9}\ C$

The charge is $3.4260289211\times 10^{-9}\ C$

The charge inside will have the polarity changed

$q=+1.7908787542\times 10^{-9}\ C$

Outside the charge will be

$3.4260289211\times 10^{-9}-1.7908787542\times 10^{-9}\\ =1.6351501669\times 10^{-9}\ C$

3 0

3 years ago

Calculate the velocity of a 1650kg satellite that is in a circular orbit of 4.2 x 10^6m above the surface of a planet which has

Anastasy [175]

-- We're going to be talking about the satellite's speed.
"Velocity" would include its direction at any instant, and
in a circular orbit, that's constantly changing.

-- The mass of the satellite makes no difference.

Since the planet's radius is 3.95 x 10⁵m and the satellite is
orbiting 4.2 x 10⁶m above the surface, the radius of the
orbital path itself is

   (3.95 x 10⁵m) + (4.2 x 10⁶m)

   = (3.95 x 10⁵m) + (42 x 10⁵m)

   = 45.95 x 10⁵ m

The circumference of the orbit is (2 π R) = 91.9 π x 10⁵ m.

The bird completes a revolution every 2.0 hours,
so its speed in orbit is

   (91.9 π x 10⁵ m) / 2 hr

= 45.95 π x 10⁵ m/hr x (1 hr / 3,600 sec)

=    0.04 x 10⁵ m/sec

      =    4 x 10³    m/sec

   (4 kilometers per second)

6 0

2 years ago