Ask question

Ask question

All categories

3 years ago

8

The area of the maxila where the teeth attach is called the ??

1 answer:

fomenos3 years ago

8 0

Answer: The alveolar process of the maxillae holds the upper teeth, and is referred to as the maxillary arch. Each maxilla attaches laterally to the zygomatic bones (cheek bones).

Explanation: wiki

You might be interested in

Green light travels by what type of wave? (2 points) a Conductive b Electronic c Conducive d Electromagnetic

antiseptic1488 [7]

Answer: d. Electromagnetic wave

Explanation:

Light is an electromagnetic wave, no matter is wavelength or color.

In this sense, the main characteristic of this type of wave is that it can propagate trhough vacumm, it is not necessary a medium. In addition, electromagnetic waves are transversal waves, this means the electric field oscillates in all normal directions to the direction of wave propagation.

8 0

3 years ago

Read 2 more answers

Compare and contrast the shape and volume of each state of matter

Inessa [10]

Answer:

<em><u>Solid</u> is the state in which matter maintains a fixed volume and shape; liquid is the state in which matter adapts to the shape of its container but varies only slightly in volume; and gas is the state in which matter expands to occupy the volume and shape of its container.</em>

7 0

2 years ago

The initial and final velocities of two blocks experiencing constant acceleration are respectively −7.45 m/s and 14.9 m/s. (a) T

ikadub [295]

Answer:

a) $a_{1}=3.7 m/s^{2}$

b) $a_{2}=3.68 m/s^{2}$

Explanation:

a) The displacement of the first object is 22.5 m, so we can use the next equation:

$v_{f}^{2}=v_{i}^{2}+2a\Delta x$

$a=\frac{v_{f}^{2}-v_{i}^{2}}{2x}$

$a=\frac{14.9^{2}-(-7.45)^{2}}{2*22.5}$

$a_{1}=3.7 m/s^{2}$

positive acceleration.

b) Using the same equation we can find the second value of the acceleration:

$a=\frac{v_{f}^{2}-v_{i}^{2}}{2x}$

$a=\frac{14.9^{2}-(-7.45)^{2}}{2*22.6}$

$a_{2}=3.68 m/s^{2}$

positive acceleration.

I hope it helps you!

8 0

3 years ago

A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the

Anvisha [2.4K]

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

$U=\dfrac{1}{2}kx^2$ ...(I)

Using gravitational potential energy

$U' =mgh$ ....(II)

Using energy of conservation

$E_{i}=E_{f}$

$U_{i}+U'_{i}=U_{f}+U'_{f}$

$\dfrac{1}{2}kx^2+0=0+mgh$

$h=\dfrac{kx^2}{2mg}$

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

$h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}$

$h=11.653\ m$

Hence, The maximum height above the point of release is 11.653 m.

3 0

3 years ago

People in the future may well live inside a rotating space structure that is more than 2 km in diameter. Inside the structure, p

Sergeeva-Olga [200]

Answer:

option B

Explanation:

given,

diameter of the rotating space = 2 Km

Force exerted at the edge of the space = 1 g

force experienced at the half way = ?

As the object is rotating in the circular part

Force is equal to centripetal acceleration.

at the edge

g = ω² r

ω is the angular velocity of the particle

r is the radius.

now, acceleration at the half way

g' = ω² r'

$g' = \omega^2 (\dfrac{r}{2})$

$g' =\dfrac{1}{2}(\omega^2 r)$

$g'=\dfrac{g}{2}$

People at the halfway experience g/2

hence, the correct answer is option B

7 0

3 years ago