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IgorC [24]
3 years ago
5

What is one layer of bone

Physics
2 answers:
devlian [24]3 years ago
8 0

Answer:

The outer surface of bone is called the periosteum (say: pare-ee-OSS-tee-um). It's a thin, dense membrane that contains nerves and blood vessels that nourish the bone. The next layer is made up of compact bone. This part is smooth and very hard.

Pie3 years ago
6 0

Periosteum. anatomy. Periosteum, dense fibrous membrane covering the surfaces of bones, consisting of an outer fibrous layer and an inner cellular layer (cambium). The outer layer is composed mostly of collagen and contains nerve fibres that cause pain when the tissue is damaged.Answer:

Explanation:

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At what distance from the wire is the magnitude of the electric field equal to 2.41 n/cn/c ?
Sladkaya [172]

The correct answer is 1.07m.

The area surrounding an electric charge where its impact may be felt is known as the electric field. When another charge enters the field, the presence of an electric field may be felt. The electric field will either attract or repel the charge depending on its makeup. Any electric charge has a property known as the electric field. The charge and electrical force working in the field determine the strength or intensity of the electric field.

Here, is the charge per unit length, r is the distance from the wire, and

is the free space permittivity  ε_0. Electric field due to the long straight wire is,

E= λ/2πε_0r

Rearrange the equation for r.

r=λ/2πε_0E

Substitute 2.41 N/C for E,

E=1.44×10^-10C/m

λ=8.85×10^-12C^2/Nm^2

r=(1.44×10^-10C/m)/(2(3.14)(8.85×10^-12C^2/Nm^2)(2.41N/C))

r=1.07m

At a distance of 1.07 m the magnitude of electric field is 2.41 N/C.

To learn more about electric field refer the link:

brainly.com/question/12821750

#SPJ4

5 0
11 months ago
An airplane awaiting departure is cleared to takeoff, its accelerates down a runway at 3.20m/ s2 for 32.8s until it finally lift
stich3 [128]

The final velocity before takeoff is 104.96 m / s.

<u>Explanation:</u>

The last velocity of a given object over some time defines the final velocity. The final velocity of the object is given by the product of acceleration and time and adding this product to the initial velocity.

To calculate the final velocity,

                          V = u + at

where v represents the final velocity,

           u represents the initial velocity,

           a represents the acceleration

            t represents the time taken.

                            v = 0 + (3.20)\times( 32.8)

                            v   = 104.96 m / s.

7 0
3 years ago
A 64.0 kg pole vaulter running at 10.2 m/s vaults over the bar. If the vaulter's horizontal component of velocity over the bar i
Lubov Fominskaja [6]

Answer:

h = 5.05 m

Explanation:

given,

mass of pole vaulter, m = 64 Kg

speed of the vaulter,V = 10.2 m/s

horizontal component of velocity in air, v = 1 m/s

height of the jump,h = ?

using energy conservation

     E_i = E_f

\dfrac{1}{2}mv_i^2 + m g h_i= \dfrac{1}{2}mv_f^2+m g h_f

initial height of the vaulter is equal to zero.

\dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2+gh_f

h =\dfrac{v_i^2-v_f^2}{2g}

h =\dfrac{10^2-1^2}{2\times 9.8}

 h = 5.05 m

height of the jump is equal to 5.05 m.

4 0
3 years ago
A soccer ball is kicked from point Pi at an angle above a horizontal field. The ball follows an ideal path before landing on the
Nutka1998 [239]

Answer:

A. The horizontal velocity vector points to the right & equals v cos θ.

Explanation:

The motion describes a parabolic path, where the horizontal speed is constant and the horizontal velocity vector always points to the right and equals v*cos θ.

8 0
3 years ago
Students use a simple pendulum with a length of 36.9cm to determine the value of "g". If it takes 14.2s to complete 10 oscillati
wel

Explanation:

It is given that,

Length of the simple pendulum, l = 36.9 cm = 0.369 m

If it takes 14.2 s to complete 10 oscillations, T=\dfrac{14.2}{10}=1.42\ s

(a) The time period of the simple pendulum is given by :

T=2\pi\sqrt{\dfrac{l}{g}}

g=\dfrac{4\pi^2l}{T^2}

g=\dfrac{4\pi^2\times 0.369}{(1.42)^2}

g=7.22\ m/s^2

(b) On the surface of moon, g'=\dfrac{g}{6}

At earth, g=9.8\ m/s^2

g'=1.63\ m/s^2

As the value of g is less on the moon, so the time period on the moon increases.

(c) The time period on the earth, T = 3 s

On earth, T=2\pi\sqrt{\dfrac{l}{g}}

3=2\pi\sqrt{\dfrac{l}{9.8}}..............(1)

On moon, T'=2\pi\sqrt{\dfrac{l}{g'}}

T'=2\pi\sqrt{\dfrac{l}{1.63}}..............(2)

On solving equation (1) and (2),

T' = 18.03 s

Hence, this is the required solution.

7 0
3 years ago
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