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3 years ago

Besides bones, the skeleton also contains connective tissues called the ???

2 answers:

7 0

Besides bones, the skeleton also contains connective tissues called the Cartilage. It is usually flexible and tough.

yan [13]3 years ago

4 0

Collagenous fibers - Purpose : Bind bones and other tissues to each other

Elastic Fibers : - Purpose : Allow organs like arteries and lungs to recoil

Reticular Fiber : - Purpose : Form a scaffolding for other cells

Hope this helps ! <3 :)

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The force of gravity acting on an object is the object's ______. A. acceleration B. mass C. weight D. matter

ryzh [129]

Answer:

Explanation:

3 0

3 years ago

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Radio waves of frequency 1.667 GHzGHz arrive at two telescopes that are connected by a computer to perform interferometry. One p

SVETLANKA909090 [29]

Explanation:

The frequency of radio waves is 1.667 GHz

One portion of the same wave front travels 1.260 mm farther than the other before the two signals are combined.

There are two conditions for interference either constructive or destructive.

For constructive interference , the path difference is n times of wavelength and for destructive interference, the path difference is (n+1/2) times of wavelength

We can find wavelength in this case as follows :

$\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{1.667\times 10^9}\\\\\lambda=0.1799\ m$

If we divide path difference by wavelength,

$\dfrac{\delta}{\lambda} =\dfrac{1.26}{0.1799}\\\\\dfrac{\delta}{\lambda} =7\\\\\delta =7\lambda$

It means that the path difference is 7 times of the wavelength. it means the two waves combine constructively and the value of m for the path difference between the two signals is 7.

5 0

3 years ago

A 6.00-μf parallel-plate capacitor has charges of 40.0 μc on its plates. how much potential energy is stored in this capacitor?

dedylja [7]

Thew energy stored in a capacitor of capacitance $C$ and voltage between the plates $V$ is

$E=\frac{1}{2} CV^2=\frac{1}{2C} Q^2$ .

Substituting numerical value

$E=\frac{1}{2*6*10^{-6}} (40*10^{-6})^2\\ E=133.33\; \mu J$

7 0

3 years ago

A vertical spring has a length of 0.25 m when a 0.175 kg mass hangs from it, and a length of 0.775 m when a 2.075 kg mass hangs

Contact [7]

Answer:

A) 35.5N/m b) 20.1cm

Explanation:

Using Hooke's law;

F = Ke where F is the weight of the object = mass of the object in kg * acceleration due to gravity in m/s^2 and k if the force constant of the spring in N/m and e is the extension of the spring which original length of the spring - new length after extension in meters

For the first body, m*g = K * (0.25- li)

Where li is the initial length of the spring

0.175*9.81 = k(0.25-li)

1.72 = k(0.25-li) as equation 1

For the second body, m *g = K* ( 0.775-li)

2.075*9.81 = k (0.775-li) equation 2

20.36 = k(0.775-li)

Make li subject of the formula;

li = 0.775 - 20.36/k

Substitute for li in equation 1

1.72 = k(0.25- (0.775 - 20.36/k))

1.72 = k ( 0.25 - 0.775 + 20.36/k)

Open the bracket with k

1.72 = 0.25k - 0.775k + 20.36 (since k cancel k)

Collect the like terms:

1.72 - 20.36 = - 0.525k

- 18.64 = -0.525k

Divide both side by -0.525

-18.64/-0.525 = -0.525/-0.525k

K = 35.5N/m

B) substitute for k in using

li = 0.775 - 20.36/k

li = 0.775 - 20.36/35.5

li = 0.775 - 0.574

li = 0.201 in meters

li = 0.201 * 100 centimeters = 20.1cm

4 0

4 years ago

Hi i need help. thNKS

Ede4ka [16]

Answer:

Explanation:

Groups go down, periods go across :)

3 0

3 years ago