A 14 pin dual-in-line IC package[14 DIL] is an integrated socket which is most popular form of IC package and has a wide range of application in digital electronics.
The 14-pin DIL has two pairs per side and each pair contains seven connecting pins.
The pairs of pins are arranged linearly one after another.The typical dimensions of width is 6.5 mm and the typical dimension of length is 18 mm.
we are asked to calculate the typical distance between two adjacent pins.
The typical distance between two adjacent pins is calculated as-
[ans]
Assuming that you have a triangular prism, the ray of light will undergo refraction twice. The first time is the transition from air to flint glass on the entry face, and the second time is the transition from the flint glass to air from the exit face. With the available data, there are two possible solution since saying "20Âş from the normal" isn't enough information. Depending upon which side of the normal that 20 degrees is, the interior triangle will have the angles of 35, 90-r, and 55+r, or 35, 90+r, 55-r degrees where r is the angle from the normal after the 1st refraction. I will provide both possible solutions and you'll need to actually select the correct one based upon the actual geometry which I don't know because you didn't provide the figure or diagram that you were provided with.
The equation for refraction is:
(sin a1)/(sin a2) = n1/n2
where
a1,a2 = angles from the normal to the surface.
n1,n2 = index of refraction for the transmission mediums.
For this problem, we've been given an a1 of 20Âş and an n1 of 1.60. For n2, we will use air which at STP has an index of refraction of 1.00029. So
(sin a1)/(sin a2) = n1/n2
(sin 20)/(sin a2) = 1.00029/1.60
0.342020143/(sin a2) = 0.62518125
0.342020143 = 0.62518125(sin a2)
0.547073578 = sin a2
asin(0.547073578) = a2
33.16647891 = a2
So the angle from the normal INSIDE the prism is 33.2Âş. The resulting angle from the surface of the entry face will be either 90-33.2 or 90+33.2 depending upon the geometry. So the 2 possible triangles will be either 35Âş, 56.8Âş, 88.2Âş or 35Âş, 123.2Âş, 21.8Âş. with a resulting angle from the normal of either 1.8Âş or 68.2Âş. I can't tell you which one is correct since you didn't tell me which side of the normal the incoming ray came from. So let's calculate both possible exits.
1.8Âş
(sin a1)/(sin a2) = n1/n2
(sin 1.8)/(sin a2) = 1.6/1.00029
0.031410759/(sin a2) = 1.599536135
0.031410759= 1.599536135(sin a2)
0.019637418= sin(a2)
asin(0.019637418) = a2
1.125213477 = a2
68.2Âş
(sin a1)/(sin a2) = n1/n2
(sin 68.2)/(sin a2) = 1.6/1.00029
0.928485827/(sin a2) = 1.599536135
0.928485827 = 1.599536135(sin a2)
0.58047193 = sin a2
asin(0.58047193) = a2
35.48374252 = a2
So if the interior triangle is acute, the answer is 1.13Âş and if the interior triangle is obtuse, the answer is 35.48Âş
Potential Energy (energy that is stored as a result of position or shape)
First, let us calculate for the volume of the block of
lead using the formula:
V = l * w * h
But we have to convert all units in terms of cm:
l = 2.0 dm = 20 cm
w = 8 cm
h = 3.5 cm
Therefore the volume is:
V = (20 cm) * (8 cm) * (3.5 cm)
V = 560 cm^3
Next we convert the mass in terms of g:
m = 6.356 kg = 6356 g
Density is mass over volume, so:
density = 6356 g / 560 cm^3
density = 11.35 g / cm^3 (ANSWER)
