<span>Lateral epicondylitis, or “tennis elbow,” is an inflammation of the tendons that join the forearm muscles on the outside of the elbow. </span>The bony bump on the outside (lateral<span> side) of the </span>elbow<span> is called the </span>lateral epicondyle<span>. The ECRB muscle and tendon is usually involved in </span>tennis elbow<span>. </span><span>
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. </span>It's identified by pain from the elbow to the wrist on the inside (medial<span> side) of the elbow. The pain is caused by damage to the tendons that bend the wrist toward the palm.</span>
Answer:
v = 5.34[m/s]
Explanation:
In order to solve this problem, we must use the theorem of work and energy conservation. This theorem tells us that the sum of the mechanical energy in the initial state plus the work on or performed by a body must be equal to the mechanical energy in the final state.
Mechanical energy is defined as the sum of energies, kinetic, potential, and elastic.
E₁ = mechanical energy at initial state [J]

In the initial state, we only have kinetic energy, potential energy is not had since the reference point is taken below 1.5[m], and the reference point is taken as potential energy equal to zero.
In the final state, you have kinetic energy and potential since the car has climbed 1.5[m] of the hill. Elastic energy is not available since there are no springs.
E₂ = mechanical energy at final state [J]

Now we can use the first statement to get the first equation:

where:
W₁₋₂ = work from the state 1 to 2.


where:
h = elevation = 1.5 [m]
g = gravity acceleration = 9.81 [m/s²]

![58 = v^{2} +29.43\\v^{2} =28.57\\v=\sqrt{28.57}\\v=5.34[m/s]](https://tex.z-dn.net/?f=58%20%3D%20v%5E%7B2%7D%20%2B29.43%5C%5Cv%5E%7B2%7D%20%3D28.57%5C%5Cv%3D%5Csqrt%7B28.57%7D%5C%5Cv%3D5.34%5Bm%2Fs%5D)
Answer
10
Explanation:
goes up by 10 each time 10 to 20 to 30
Answer:
E = k Q / [d(d+L)]
Explanation:
As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field
E = k ∫ dq/ r² r^
"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.
Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant
λ = Q / L
If we derive from the length we have
λ = dq/dx ⇒ dq = L dx
We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge
dE = k dq / x²2
dE = k λ dx / x²
Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider
E = k 
We take out the constant magnitudes and perform the integral
E = k λ (-1/x)
Evaluating
E = k λ [ 1/d - 1/ (d+L)]
Using λ = Q/L
E = k Q/L [ 1/d - 1/ (d+L)]
let's use a bit of arithmetic to simplify the expression
[ 1/d - 1/ (d+L)] = L /[d(d+L)]
The final result is
E = k Q / [d(d+L)]
fraction equation is<span>
F =µR
F=friction,µ=coefficient , R=reaction = mg
use same equation for b part, but the reaction is no longer mg because the plain is now inclined. Draw a forces diagram and you will see that the reaction force can be calculated from the weight of the object and inclination of the plain using trigonometry.</span>