Answer:
(E) μs(mA +mB)g
Explanation:
We can apply for mB:
∑ Fx = mB*a (→)
⇒ Ffriction = mB*a ⇒ a = Ffriction / mB = μs*N / mB
⇒ a = μs*(mB*g) / mB ⇒ a = μs*g (acceleration of the system)
Now, for mA we have
∑ Fx = mA*a (→)
F - Ffriction = mA*a ⇒ F = mA*a + Ffriction
⇒ F = mA*(μs*g) + μs*(mB*g) ⇒ F = μs*g*(mA + mB)
We must know that the friction acts only between the two blocks
Assume that an ingot of copper has a mass of 9.1 kg or 9100 g.
The cross-sectional area of the copper wire with diameter of 6.5 mm (or 0.65 cm) is
A = (π/4)*(0.65 cm)² = 0.3318 cm²
The density of copper is given as 8.94 g/cm³.
If the length of copper wire is L cm, then
(0.3318 cm²)*(L cm)*(8.94 g/cm³) = 9100 g
L = 9100/(0.3318*8.94) = 3.0678 x 10³ cm
Note that
1 cm = 1/2.54 in = 1/2.54 in = 0.3937 in
= 0.3937/12 = 0.03281 ft
Therefore
L = (3.0678 x 10³ cm)*(0.03281 ft/cm) = 100.65 ft
Answer: 100.65 ft
Answer:
1. b
2. d
3. e
4. c
5. a
Explanation:
These are just basic definitions. Let me know if you need further clarification.
