How does the elbow medical and lateral epicondylitis?

1 answer:

4 0

Lateral epicondylitis, or “tennis elbow,” is an inflammation of the tendons that join the forearm muscles on the outside of the elbow. The bony bump on the outside (lateral side) of the elbow is called the lateral epicondyle. The ECRB muscle and tendon is usually involved in tennis elbow. 
Medial epicondylitis, or “golfer’s elbow,” is an inflammation of the tendons that attach your forearm muscles to the inside of the bone at your elbow. It's identified by pain from the elbow to the wrist on the inside (medial side) of the elbow. The pain is caused by damage to the tendons that bend the wrist toward the palm.

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Kipish [7]

Answer:

The intensity of the electric field is

$|E|=10654.37 \:N/C$

Explanation:

The electric field equation is given by:

$|E|=k\frac{q}{d^{2}}$

Where:

k is the Coulomb constant
q is the charge at 0.4100 m from the balloon
d is the distance from the charge to the balloon

As we need to find the electric field at the location of the balloon, we just need the charge equal to 1.99*10⁻⁷ C.

Then, let's use the equation written above.

$|E|=(9*10^{9})\frac{1.99*10^{-7}}{0.41^{2}}$

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I hope it helps you!

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Elements in an blank or blank of the periodic table have similar characteristics

mr_godi [17]

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Underground water is being pumped into a pool whose cross section is 3 m x 4 m while water is discharged through a 0.076m-diamet

Svetllana [295]

Given:

Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min

Required:

Inlet flowrate

Solution:

The problem can be solved by this general formula.

Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

First, we need to convert the units of the accumulation velocity into m/s to be consistent.

Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s

We then calculate the area of the pool and the area of the orifice by:

Area of pool = 3 × 4 m²
Area of pool = 12m²

Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²

Since we have all we need, we plug in the values to the general equation earlier

Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²

Transposing terms,

Inlet flowrate = 0.316 m³/s

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