A street lamp weighs 150N. It is supported by two wires that form an angle of 120° with each other. The tensions in each wire ar
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Answer:
The resultant tension of the two ropes is approximately 42.4 N
The length of the line representing the resultant tension is approximately 8.48 cm
Please find included with the answer the scale drawing created with Microsoft Word
Explanation:
The given parameters are;
The tension in rope P, = 30 N
The tension in rope Q, = 30 N
The angle the rope, 'P', makes with the horizontal = 45°
The angle the rope, 'Q', makes with the horizontal = 45°
The scale factor of the scale diagram, S.F. = 5.0 N/cm
By the resolution of forces at equilibrium, we have;
The sum of the vertical forces, =
+
+ W = 0
∴ W = -( +
)
W = -(30 × sin(45°) + 30 × sin(45°)) = -42.4264068712
The weight of the heavy ball, W ≈ 42.4 N acting downwards
The sum of the horizontal forces, =
+
= 0
The length of the resultant force, W = W/(S.F.) ≈ 42.4 N/(5.0 N/cm) = 8.48 cm
The drawing of the vectors using the scale factor of 5.0 N/cm is created using Microsoft Word is included
In physical chemistry, the terms body-centered cubic (BCC) and face-centered cubic (FCC) refer to the cubic crystal system of a solid. Each solid is made up simple building blocks called lattice units. There are different layouts of a lattice unit.
It is better understood using 3-D models shown in the picture. A BCC unit cell has one lattice point in the center, together with eight corner atoms which represents 1/8 of an atom. Therefore, there are 1+ 8(1/8) = 2 atoms in a BCC unit cell. On the other hand, a FCC unit cell is composed of half of an atom in each of its faces and 1/8 of an atom in its corners. Therefore, there are (1/2)6 + (1/8)8 = 4 atoms in a FCC unit cell.
It is better understood using 3-D models shown in the picture. A BCC unit cell has one lattice point in the center, together with eight corner atoms which represents 1/8 of an atom. Therefore, there are 1+ 8(1/8) = 2 atoms in a BCC unit cell. On the other hand, a FCC unit cell is composed of half of an atom in each of its faces and 1/8 of an atom in its corners. Therefore, there are (1/2)6 + (1/8)8 = 4 atoms in a FCC unit cell.