Answer:
a. Fab = 3000 N tension
Fac = 1500 N compression
Fbc = 3000 N compression
b. Aluminum for AB, titanium for AC and BC.
c. Fc = 4000 N
Explanation:
a. Draw free body diagrams of the pins at A and B, assuming that all internal forces are in tension (pulling away from the pin).
Sum of the forces on A in the x direction:
∑F = ma
H − Fab sin 60° = 0
Fab = H / sin 60°
Fab = 2600 / sin 60°
Fab = 3000 N
Sum of the forces on A in the y direction:
∑F = ma
-Fab cos 60° − Fac = 0
Fac = -Fab cos 60°
Fac = -H / tan 60°
Fac = -2600 / tan 60°
Fac = -1500 N (it's negative, so it's actually in compression).
Sum of the forces on B in the x direction:
∑F = ma
Fab cos 30° + Fbc cos 30° = 0
Fbc = -Fab
Fbc = -H / sin 60°
Fbc = -3000 N (it's negative, so it's in compression)
b. AB is in tension, so it should be aluminum.
AC and BC are in compression, so they should be titanium.
c. Draw a free body diagram of the entire triangle ABC. Assume there is both a horizontal and vertical reaction force at C, and assume they point in the +x and +y directions.
Sum of the forces in the x direction:
∑F = ma
Fcx + H = 0
Fcx = -H
Fcx = -2600 N
Sum of the forces in the y direction:
∑F = ma
Fcy − V = 0
Fcy = V
Fcy = 3000 N
The magnitude of the net force is:
Fc² = Fcx² + Fcy²
Fc² = (-2600 N)² + (3000 N)²
Fc = 4000 N
Answer:
(4) more protons than electrons
Explanation:
Total positive change on the object is given as
now by the law of quantization of charge we know that the charge on any body is always in form of integral multiple of charge of an electron.
So we will have
here we know that
Now we know that the object is positively charged so here electrons must be less in numbers than protons
so correct answer will be
(4) more protons than electrons