All categories

4 years ago

What is an example of a partially movable joint?

1 answer:

3 0

<span>A partially movable joint is a joint in which its movement is limited to a certain amount. They are usually supported by a layer of cartilages and fibrous connective tissues. They are also called amphiarthrosis joints. Example of this are the joints formed by adjacent vertebrae containing intervertebral disc of a cartilage. When the joints are moved, it limits its movement because of this arrangement. Other examples of this are the ribs and the sternum. Notice that you cannot twist your upper body at 360 degrees without moving your lower body.</span>

You might be interested in

33 POINTS How can I get the temperature? SOUND SPEED 340 m/s = 331 m/s + (0,6xTemperature)

inessss [21]

(340-331)/0.6 = temp

9/0.6

90/6

30/2

15 degrees

6 0

3 years ago

A solid sphere of brass (bulk modulus of 14.0 ✕ 1010 N/m2) with a diameter of 2.20 m is thrown into the ocean. By how much does

astra-53 [7]

Answer:

Diameter decreases by the diameter of 0.0312 m.

Explanation:

Given that,

Bulk modulus = 14.0 × 10¹⁰ N/m²

Diameter d = 2.20 m

Depth = 2.40 km

Pressure = ρ g h = 1030 × 9.81 × 2.4 × 1000

= 24.25 × 10⁶ N/m²

Volume = $\dfrac{4}{3} \pi r^3$

$\dfrac{\Delta V}{V}=\dfrac{(\Delta r)^3}{r^3}$

Bulk modulus is equal to

$B = -\dfrac{\Delta P}{\dfrac{\Delta V}{V} }$

now

$B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{r^3} }$

$B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{1.1^3} }$

$(\Delta r)^3 = \dfrac{24.25 \times 10^6 \times 1.1^3}{14.0 \times 10^{10}}$

Δ r = -0.0156 m

change in diameter

Δ d = -2 × 0.0156

Δ d = -0.0312 m

Diameter decreases by the diameter of 0.0312 m.

7 0

3 years ago

Simplified. The absorption spectra of ions have been used to identify the presence of the elements in the atmospheres of the sun

Nookie1986 [14]

Answer:

The answer is " $3.83 \times 10^9 \ m$ "

Explanation:

Z=2, so the equation is $E= \frac{-4B}{n^2}$

Calculate the value for E when:

n=2 and n=9

The energy is the difference in transformation, name the energy delta E Deduct these two energies

In this transition, the wavelength of the photon emitted is:

$\Delta E=2.18 \times 10^{-18} ( \frac{1}{4}- \frac{1}{81})$

$\lambda = \frac{h c}{\Delta E}$

$h ( Planck's\ constant) = 6.62 \times 10^{(-34)} \ Js \\\\ speed \ of \ light = 3 \times 10^{8} \ \frac{m}{s}\\\\= \frac{6.62 \times 10^{(-34)} \times 3 \times 10^ {8}}{2.18 \times 10^{-18}} (\frac{1}{4}- \frac{1}{81}) \\\\=3.83 \times 10^9 \ m\\\\$

6 0

3 years ago

The formula for calculating work is ___.

Orlov [11]

Hey!

Formula for calculating work = Force × distance
Work done = Force applied on the object × distance travelled by it.

Hope it helps...!!!

3 0

4 years ago

A particle moves in a velocity field V(x, y) = x2, x + y2 . If it is at position (x, y) = (7, 2) at time t = 3, estimate its loc

MArishka [77]

Answer:

New location at time 3.01 is given by: (7.49, 2.11)

Explanation:

Let's start by understanding what is the particle's velocity (in component form) in that velocity field at time 3:

$V_x=x^2=7^2=49\\V_y=x+y^2=7+2^2=11$

With such velocities in the x direction and in the y-direction respectively, we can find the displacement in x and y at a time 0.01 units later by using the formula:

$distance=v\,*\, t$

$distance_x=49\,(0.01)=0.49\\distance_y=11\,(0.01)=0.11$

Therefore, adding these displacements in component form to the original particle's position, we get:

New position: (7 + 0.49, 2 + 0.11) = (7.49, 2.11)

6 0

3 years ago