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tino4ka555 [31]
3 years ago
7

Bursitis can affect many joints and can be caused by a numerous sporting activities, such as cycling, tennis, and long-distance

running. Explain what bursitis is and what characteristics the mentioned sporting activities have in common?
Physics
2 answers:
Natali5045456 [20]3 years ago
8 0

Bursitis is a painful condition that affects the joints. Bursae are fluid-filled sacs that act as a cushion between bones, tendons, joints, and muscles. When these sacs become inflamed it is called bursitis. 

Some causes of bursitis include:

Tennis elbow: Bursitis is a common problem among tennis players and golfers. Repetitive bending of the elbow can lead to injury and inflammation.

Clergyman's knee: Repeated kneeling can cause injury and swelling to the bursae in the knee area.

Shoulder: Repeated overhead lifting or reaching upwards can cause bursitis in the shoulder.

Ankle: Injury to the ankle can result from walking too much and with the wrong shoes. It is common among ice skaters and athletes.

Buttocks: The bursae in this area can become inflamed after sitting on a hard surface for a long time, such as on a bicycle.

Hips: Some runners and sprinters can develop hip bursitis.

Thigh: Bursitis can be caused by stretching.

DerKrebs [107]3 years ago
6 0
Bursitis is when the small fluid filled sacs around the joints called bursae are irritated or injured. In just about any joint the effects can be felt with pain in movement and when pressed upon. Sometimes a rash will occurs as well. Thank you for posting your question here. I hope the answer will help. <span>
</span>
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Calculate the potential energy of a rock that has a 45 kg mass and is sitting
Montano1993 [528]

Answer:

P.E = 13230 J

Explanation:

Given,

The mass of the rock, m = 45 Kg

The rock is sitting at a height from the ground, h = 30 m

The acceleration due to gravity, g = 9.8 m/s²

The potential energy of the body is given by the formula,

                            P.E = mgh joules

Substituting the given values in the above equation

                             P.E = 45 x 9.8 x 30

                                   = 13230 J

Hence, the potential energy of the rock is P.E = 13230 J

6 0
3 years ago
What is the magnitude of the net force acting on a 2.0 x10^3 kilogram car?
Tatiana [17]
Answer:  
Acceleration= final velocity - initial velocity / time
 = (15 - 0)/ 5
 = 3 m/s^2
 Force = mass X acceleration
 = 2 X 10^3 X 3
 = 6 X 10^3 N
7 0
3 years ago
Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.8×106N , one an angle 11degrees west of north an
sp2606 [1]

Answer: 2,9 ×10¹⁰J.

Explanation:

1. The <em><u>work </u></em>is a measure of energy and is calculated as the product of the displacement times the parallel force to such displacement.

That means that the only components of the force that contribute to work are those that result parallel to the displacement.

2. Since it is given that the <em>two tugboats "pull the tanker a distance 0.83km toward the north"</em>, that is the displacement, and you have to calculate the net force toward the north.

3. <u>Tugboat #1</u>.

a) Force magnitude: F₁ = 1.8×10⁶N

b) Angle: α = 11° West of North

c) North component of the force F₁: Fy₁ = F₁cos(α) =  1.8×10⁶N  × cos(11°) = 1.77×10⁶N

4. <u>Tugboat #2</u>:

a) Force magnitude: F₂ = 1.8×10⁶N

b) Angle:  = 11° East of North

c) North component of the force F₂: Fy₂ = F₂cos(β) =  1.8×10⁶N  × cos(11°) = 1.77×10⁶N =

5. <u>Total net force, Fn</u>:

Fn = Fy₁ + Fy₂ = 1.77×10⁶N + 1.77×10⁶N = 3.54×10⁶N

6. <u>Work, W</u>:

Displacement, d = 0.83 km = 8,300 m

W = Fn×d = 3.54×10⁶N×8,300m = 29,000 ×10⁶J = 2,9 ×10¹⁰J

The answer is rounded to two significant figures because both data, Force and displacement, have two significant figures.

7 0
3 years ago
Is Experience Nature or Nurture
Stella [2.4K]

Answer:

Nature

Explanation:

6 0
4 years ago
Please solve the Problem.
erica [24]
  • The equivalent capacitance between point a and b is 5.87μf.
  • The charge at 20μf is 93.92 μC.
  • The charge at 6μf is 67.8 μC.
  • The charge at C and 3μf is 26.12 μC.

<h3>Sum of capacitance of C and 3μF</h3>

The sum of the capacitance is calculated as follows;

1/Ct = 1/C + 1/3

1/Ct = 1/10μf + 1/3μf

1/Ct = (3μf + 10μf)/30μf²

1/Ct = 13μf/30μf²

Ct = 30μf²/13μf

Ct = 2.31μf

<h3>Total capacitance in parallel arrangement</h3>

The total capacitance in parallel arrangement is calculated as follows;

Ct = 2.31μf + 6μf = 8.31μf

<h3>Equivalent capacitance between point a and b</h3>

1/Ct = 1/8.31μf + 1/20μf

1/Ct = 0.1703

Ct = 5.87μf

<h3>Charge flowing in each capacitor</h3>

Maximum voltage is delivered in 20μf, q = CV

<u>charge for 20μf:</u>

q = (5.87 x 16)μC

q = 93.92 μC

<h3>Equivalent capacitance for C, 3μf and 6μf</h3>

Ct = 2.31μf + 6μf = 8.31uf

<u>charge for 6μf:</u>

q = (6/8.31) x 93.92μC

q = 67.8μC

<h3>Total charge for C and 3μf</h3>

q = 93.92μC - 67.8μC = 26.12 μC

charge for C = charge 3μf = 26.12 μC

Learn more about capacitor here: brainly.com/question/14883923

#SPJ1

3 0
3 years ago
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