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artcher
3 months 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:
serg [3.5K]3 months ago
8 0

Bursitis is characterized by discomfort in the joints. The bursae are the small sacs filled with fluid, providing cushioning between bones, tendons, joints, and muscles. When these bursae become inflamed, it is termed bursitis.

Common causes of bursitis include:

Tennis elbow: This condition is prevalent among golfers and tennis players, where repetitive elbow bending leads to inflammation and injury.

Clergyman's knee: Persistent kneeling can cause swelling and injury to the bursae around the knee.

Shoulder: Consistent overhead lifting or reaching can result in bursitis in the shoulder region.

Ankle: Walking excessively or wearing inappropriate footwear can cause ankle issues, often seen in ice skaters and athletes.

Buttocks: Prolonged sitting on hard surfaces, like riding a bicycle, may irritate the bursae in this area.

Hips: Hip bursitis can be common among runners and sprinters.

Thigh: Stretching may result in bursitis as well.

Yuliya22 [3.3K]3 months ago
6 0
Bursitis refers to the irritation or injury of the small fluid-filled sacs called bursae that are located around the joints. This condition can lead to pain during movement or when the affected area is pressed. Occasionally, a rash may also develop. Thank you for your inquiry; I trust this information will be beneficial.
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An 8.0 m, 240 N uniform ladder rests against a smooth wall. The coefficient of static friction between the ladder and the ground
Ostrovityanka [3204]

Answer:

5.7 m

Explanation:

AD = length of the ladder = L = 8 m

AB = the position of the ladder's center of mass = (0.5) L = (0.5) 8 = 4 m

AC = distance of the climber from the bottom of the ladder = x

W = weight of the ladder = 240 N

F_{g} = weight of the climber = 710 N

F = force exerted by the wall on the ladder

N = normal force acting on the ladder from the ground =?

By applying force equilibrium in the vertical direction

N = F_{g} + W

N = 710 + 240

N = 950 N

μ = Coefficient of static friction = 0.55

f = static friction force on the ladder

Static friction force can be expressed as

f = μ N

f = (0.55) (950)

f = 522.5 N

The equation for force along the horizontal axis reads

F = f

F = 522.5 N

using torque equilibrium around point A

F Sin50 (AD) = W Cos50 (AB) + (F_{g} Cos50 (AC))

(522.5) Sin50 (8) = (240) Cos50 (4) + (710) Cos50 (x)

x = 5.7 m

7 0
2 months ago
Now slowly begin to raise the temperature. At approximately what temperature would a heated material (metal, wood, etc.) begin t
ValentinkaMS [3465]
We will utilize Wien's displacement law, given by the equation λ T = b, where λ represents the wavelength of emitted light from a heated object at maximum. By substituting the provided temperature and constant b into the equation, we find λ for various temperatures: at 500 K, λ = 5.796 μm or 5796 nm; at 1050 K, λ = 2760 nm; at 1800 K, λ = 1610 nm; and at 2500 K, λ = 1159.2 nm. The visible light spectrum starts at 740 nm, suggesting that at 2500 K, some visible red light may emerge as its calculated peak wavelength is within the visible range.
3 0
2 months ago
A hot air balloon of total mass M (including passengers and luggage) is moving with a downward acceleration of magnitude a. As i
inna [3103]

Answer:

The ratio of mass that is discarded is determined by this equation:

M - m = (3a/2)/(g²- (a²/2) - (ag/2))

Explanation:

The force acting on an object in motion is defined by the equation:

F = ma

Additionally, there is a gravitational force consistently acting downwards on the object, defined as g = 9.8 ms⁻²

For convenience, we will utilize a positive notation for downward acceleration and a negative notation for upward acceleration.

Case 1:

The hot air balloon has mass = M

Acceleration = a

Upward thrust from hot air = F = constant

Gravitational force acting downward = Mg

The net force on the balloon can be expressed as:

Ma = Gravitational force - Upward Force                              

Ma = Mg - F                      (since the balloon moves downward, that means Mg > F)

F = Mg - Ma

F = M (g-a)

M = F/(g-a)

Case 2:

After releasing the ballast, the new mass becomes m. The new upward acceleration is -a/2:

The net force is expressed as:

-m(a/2) = mg - F        (The balloon is moving upwards, hence F > mg)

F = mg + m(a/2)

F = m(g + (a/2))

m = F/(g + (a/2))

Determining the fraction of the mass initially dropped:

M-m = \frac{F}{g-a} - \frac{F}{g+\frac{a}{2} }\\M-m = F*[\frac{1}{g-a} - \frac{1}{g+\frac{a}{2} }]\\M-m = F*[\frac{(g+(a/2)) - (g-a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{g+(a/2) - g + a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{(3a/2)}{g^{2}-\frac{a^{2}}{2}-\frac{ag}{2}} ]

5 0
3 months ago
A 25kg child sits on one end of a 2m see saw. How far from the pivot point should a rock of 50kg be placed on the other side of
Sav [3153]

Answer:

A rock weighing 50kg should be positioned at a distance of 0.5m from the pivot of the seesaw.

Explanation:

τchild=τrock  

We will utilize the formula for torque:

(F)child(d)child)=(F)rock(d)rock)

The gravitational force acts equally on both objects.

(m)childg(d)child)=(m)rockg(d)rock)

We can eliminate gravity from both sides of the equation for simplification.

 (m)child(d)child)=(m)rock(d)rock)  

Now employing the given masses for the rock and child. The seesaw's total length is 2 meters, with the child sitting at one end, placing them 1 meter from the center of the seesaw.

(25kg)(1m)=(50kg)drock

Solve for the distance where the rock should be positioned in relation to the seesaw's center.

drock=25kg⋅m50kg

drock=0.5m

6 0
3 months ago
A car rolls down a ramp in a parking garage. The horizontal position of the car in meters over time is shown below. Graph of ver
Ostrovityanka [3204]

Answer:

d_total = 12 m

Explanation:

In this kinematics scenario illustrated in the graph provided, we determine the distance traveled over a 24-second duration.

The comprehensive distance can be calculated as follows:

d_total = d₁ + d₂ + d₃

Given that d₂ on the graph is level (v=0), its distance equates to zero, hence d₂ = 0.

The distance for d₁ is calculated as:

d₁ = 12 - 6 = 6 m

For distance d₃:

d₃ = 6 - 0 = 6 m

Thus, the overall distance covered is:

d_total = 6 + 0 + 6

d_total = 12 m

3 0
3 months ago
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