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artcher
1 month 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]1 month 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]1 month 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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A 1.97-pF capacitor with a plate area of 5.86 cm2 and separation between the plates of 2.63 mm is connected to a 9.0-V battery a
inna [3103]

If the plate separation is modified after the battery is disconnected, the updated distance between plates is 9.21 mm

If changes are made while the battery remains connected, the new separation becomes 0.11 mm

The capacitance for an air-filled parallel plate capacitor can be expressed as:

C=\frac{\epsilon_0A}{d}

In this equation, \epsilon_0 refers to the permittivity of free space, A stands for the plate area, and D represents the separation distance.

Thus,

C \alpha \frac{1}{d}.......(1)

Therefore, should the distance between the plates shift from d₁ to d₂, the capacitance ratio in both scenarios can be represented as:

\frac{C_1}{C_2} =\frac{d_2}{d_1}......(2)

Scenario (i)

When the capacitor is fully charged and then disconnected from the battery before adjusting the plate distance, the charge will remain steady while the capacitance varies.

The initial energy E₁ stored in the capacitor can be expressed as:

E_1=\frac{Q^2}{2C_1}......(3)

Once the separation changes to d₂, capacitance becomes C₂, but the charge Q remains unchanged.

Thus,

E_2=\frac{Q^2}{2C_2}......(4)

By dividing equation (4) by (3),

\frac{E_2}{E_1} =\frac{C_1}{C_2}

According to equation (2),

\frac{E_2}{E_1} =\frac{C_1}{C_2}=\frac{d_2}{d_1}

This results in a 3.5 fold increase in energy.

\frac{E_2}{E_1} =\frac{d_2}{d_1}=3.5\\ d_2=3.5*2.63 mm\\ =9.205 mm=9.21 mm

Scenario (2)

If the capacitor is kept connected to the power source, the voltage V across the plates will remain unchanged.

The initial energy is described as

E_1=\frac{1}{2} C_1V^2......(5)

The final energy when the plate separation transitions to d₂ can be written as:

E_2=\frac{1}{2} C_2V^2.....(6)

Referencing equations (5) and (6)

\frac{E_2}{E_1} =\frac{C_2}{C_1}

From equation (2),

\frac{E_2}{E_1} =\frac{C_2}{C_1}=\frac{d_1}{d_2}

Thus, in this particular scenario,

\frac{E_2}{E_1} =\frac{d_1}{d_2}\\d_2=\frac{d_1}{3.5} \\ =\frac{2.63 mm}{3.5} \\ =0.109 mm=0.11 mm

Therefore,

Adjusting plate separation after battery disconnection yields 9.21 mm

If modified while connected, the new separation measures 0.11 mm





6 0
1 month ago
If the coefficient of static friction between a table and a uniform massive rope is μs, what fraction of the rope can hang over
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Here's the procedure explained: Assume F represents the portion of the rope that is extending over the table. In this scenario, the frictional force that holds the rope on the table can be calculated using the formula: Ff = u*(1-f)*m*g. Additionally, it is important to determine the gravitational force that attempts to pull the rope off the table, Fg, calculated through: Fg = f*m*g. You then need to set these two equations equal to each other and resolve for f: f*m*g = u*(1-f)*m*g leads to f = u*(1-f) = u - uf. Simplifying gives f + uf = u, which results in f = u/(1+u) representing the fraction of the rope. This will lead you to the final answer.
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1 month ago
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At a given point on a horizontal streamline in flowing air, the static pressure is â2.0 psi (i.e., a vacuum) and the velocity is
Softa [3030]
Bernoulli's equation at a point on the streamline is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = air density, 0.075 lb/ft³ (under standard conditions)
g = 32 ft/s²

Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s

Point 2 (stagnation):
The velocity at the stagnation point is zero.

The density stays constant.
Let p₂ denote the pressure at the stagnation location.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
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27 days ago
Which pair of graphs represent the same motion of an object
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The correct choice is C
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2 months ago
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Which of the following combinations of variables results in the greatest period for a pendulum? length = L, mass = M, and maximu
Yuliya22 [3333]

Response:

length = 2L, mass = M/2, and maximum angular displacement = 1 degree

Clarification:

We examine only small amplitude oscillations (as in this scenario), which keeps the angle θ sufficiently small. In such situations, it's important to note that the pendulum's motion can be described by the equation:

\ddot{\theta}=\frac{g}{l}\theta

The resulting solution is:

\theta=Asin(\omega t + \phi)

Here, \omega=\sqrt\frac{g}{l} represents the angular frequency of the oscillations, enabling us to find the period:

T=\frac{2\pi}{\omega}\\T=2\pi\sqrt\frac{l}{g}

As a result, the period of a pendulum is determined solely by its length and is independent of both its mass and angle, provided the angle remains small. Therefore, the choice with the longest length gives the longest period.

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