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3 years ago

The ________ on the axis (c2) forms a pivot point with the atlas (c1) that allows you to nod a "no."

1 answer:

3 0

The dens or the odontoid process of the axis or the second cervical spine forms a pivot point with the atlas or the first cervical vertebrae that is responsible for the nodding and the rotational movements of the head. This is reinforced by ligaments and the atlanto-occipital joint that allows the head to make a nodding or up and down movement on the vertebral column.

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Tarik winds a small paper tube uniformly with 183 turns 183 turns of thin wire to form a solenoid. The tube's diameter is 9.49 m

Rufina [12.5K]

<h2>Answer:</h2>

143μH

<h2>Explanation:</h2>

The inductance (L) of a coil wire (e.g solenoid) is given by;

L = μ₀N²A / l --------------(i)

Where;

l = the length of the solenoid

A = cross-sectional area of the solenoid

N= number of turns of the solenoid

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

<em>From the question;</em>

N = 183 turns

l = 2.09cm = 0.0209m

diameter, d = 9.49mm = 0.00949m

A = π d² / 4 [Take π = 3.142 and substitute d = 0.00949m]

A = 3.142 x 0.00949² / 4

A = 7.1 x 10⁻⁵m²

<em>Substitute these values into equation (i) as follows;</em>

L = 4π x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209 [Take π = 3.142]

L = 4(3.142) x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209

L = 143 x 10⁻⁶ H

L = 143 μH

Therefore the inductance in microhenrys of the Tarik's solenoid is 143

6 0

2 years ago

The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, th

Inga [223]

According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

$E_i = E_f$

$0 = \frac{1}{2} (m_1+m_2)v_f^2-m_2gh+m_1gh$

$v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}$

PART B) Replacing the values given as,

$h= 1.7m\\m_1 = 3.5kg\\m_2 = 4.3kg \\g = 9.8m/s^2 \\$

$v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}$

$v_f = \sqrt{2(9.8)(1.7)(\frac{4.3-3.5}{3.5+4.3})}$

$v_f = 1.8486m/s$

Therefore the speed of the masses would be 1.8486m/s

6 0

3 years ago

Pressure of gas is 13.6 cm of hg what does it means

vlada-n [284]

Explanation:

13 cmHg (centimeters of mercury) is the pressure at the bottom of a column of mercury 13 cm deep. It is the equivalent of about 17.3 kPa or 2.5 psi.

5 0

3 years ago

Two asteroids identical to those above collide at right angles and stick together; i.e, their initial velocities were perpendicu

11111nata11111 [884]

Answer:

velocity = 62.89 m/s in 58 degree measured from the x-axis

Explanation:

Relevant information:

Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.

Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $10^5$ kg - m/s in the right direction.