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In-s [12.5K]
3 years ago
8

Identify break down and define 10 medical terms related to the musculoskeletal system

Physics
1 answer:
natita [175]3 years ago
7 0
1. Tendinitis
Tendin/ - refers to the tendon
/itis - means inflammation

Tendinitis is the inflammation of the tendon. It is usually caused by repetitive impact on the area that is affected and/or direct injury. 

2. Osteoporosis
Oste/o/ - refers to bones 
/porosis - means porous or a condition of pores.

Osteoporosis literally means porous bones. It is a disease where bones become fragile or weak because of the decreased bone density. It increases the risk of fracture. 

3. Arthritis
Arthr(o)/ - refers to joints
-itis - means inflammation

Arthritis is the inflammation of a joint or joints. Joints affected develop stiffness, pain, redness, and warmth. It can be caused by infection, metabolic problems and or constitutional causes. 

4. Osteocytes
Osteo/ - refers to the bone
cyte/s - Means cells. 

Osteocytes are also known as bone cells. It is the type of cell that can be found in bones that are fully formed. 

5. Costochondritis
Cost(o)/ - refers to the ribs.
chondr(o)/ - refers to cartilage
-itis - means inflammation

Costochondritis is the inflammation of the cartilage connected to the sternum and the ribs. It is also known as chest wall pain and the pain resembles the pain felt during a heart attack. 

6. Lordosis
Lord(o)/ -means curve/swayback
-osis - pertains to an abnormal condition

Lordosis is also known as swayback. This occurs when the spine curve is too far inward in the lumbar spine.

7. Myocytes
Myo/ -refers to the muscles
cyte/s - means cells. 

Myocytes are also known as muscle cells. They are the type of cells that make up tissues of the muscles. They are long and tubular in appearance. 

8. Myelocyte
Myelo/ -refers to the bone marrow or the spinal cord.
/cyte - means cell.

Myelocytes are bone marrow cells. They are motile and produce blood granulocytes (these are a kind of white blood cells). They occur in the bone marrow. 

9. Calcaneodynia
Calcane/o - refers to the heel bone
/dynia - pertains to pain

Calcaneodynia is a condition where there is pain when weight is put on the heel of the foot. 

10. Metacarpectomy
Metacarp(o)/ - refers to bones in the hand (Metacarpals) 
/ectomy - excision or removal.

Metacarpectomy is the surgical excision of one or more metacarpals. 



 
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Average speed equals distance / time

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The resistance, R, of a wire varies directly as its length and inversely as the square of its diameter. If the resistance of a w
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R is proportional to the length of the wire:

R ∝ length

R is also proportional to the inverse square of the diameter:

R ∝ 1/diameter²

The resistance of a wire 2700ft long with a diameter of 0.26in is 9850Ω. Now let's change the shape of the wire, adding and subtracting material as we go along, such that the wire is now 2800ft and has a diameter of 0.1in.

Calculate the scale factor due to the changed length:

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Scale factor due to changed diameter:

k₂ = 1/(0.1/0.26)² = 6.76

Multiply the original resistance by these factors to get the new resistance:

R = R₀k₁k₂

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8 0
3 years ago
In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the
Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, u_0= 10 m/s.

Angle of launch with the horizontal, \theta = 53 ^{\circ}

So, the vertical component of the initial velocity,

u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i).

The horizontal component of the initial velocity,

u_0\cos\theta=10 \cos 53 ^{\circ}

Let, t be the time of flight, to the horizontal distance covered

D=10 \cos (53 ^{\circ})t\cdots(ii).

Not, applying the equation of motion in the vertical direction.

s= ut +\frac 1 2 at^2

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, u =10 \sin 53 ^{\circ} (from equation (i), s=0 (as the final height is same as the launch height) and a = -9.81 m/s^2 (negative sign is due to the downward direction).

\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2

\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63 seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m.

Now, for the maximum height, H, applying the equation of motion as

v^2=u^2+2as

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, 0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H

\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}

\Rightarrow H = 3.25 m.

Hence, the maximum height covered is 3.25 m.

8 0
3 years ago
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