<span>when two hydrogen atoms approach each other the potential energy of the combination becomes lower and lower until it reaches a minimum value of -436 kl/mol at a distance of 75 pm</span>
I believe the answer is heat.
To solve this problem it is necessary to apply the concepts related to uniaxial deflection for which the training variable is applied, determined as
Where,
P = Tensile Force
L= Length
A = Cross sectional Area
E = Young's modulus
PART A) The elongation of the bar in a length of 200 mm caliber, could be determined through the previous equation, then
Therefore the elongaton of the rod in a 200mm gage length is 
PART B) To know the change in the diameter, we apply the similar ratio of the change in length for which,
Where,
Poission's ratio
= Lateral strain
= Linear strain
Therefore the change in diameter of the rod is 
Answer: discplacement/change in time = average velocity.
Explanation:
There is a formula for velocity which is given above
Answer:
19.6 cm.
Explanation:
From the question given above, the following data were obtained:
Focal length (f) = 13.8 cm
Magnification (M) = +2.37
Object distance (u) =.?
Next, we shall determine the image distance. This can be obtained as follow:
Magnification (M) = +2.37
Object distance (u) = u
Image distance (v) =?
M = v / u
2.37 = v / u
Cross multiply
v = 2.37 × u
v = 2.37u
Finally, we shall determine the object distance. This can be obtained as follow:
Focal length (f) = 13.8 cm
Image distance (v) = 2.37u
Object distance (u) =.?
1/v + 1/u = 1/f
vu / v + u = f
2.37u × u / 2.37u + u = 13.8
2.37u² / 3.37u = 13.8
Cross multiply
2.37u² = 3.37u × 13.8
2.37u² = 46.506u
Divide both side by u
2.37u² / u = 46.506u / u
2.37u = 46.506
Divide both side by 2.37
u = 46.506 / 2.37
u = 19.6 cm
Thus, the lens should be held at a distance of 19.6 cm.
