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:
114.44 J
Explanation:
From Hook's Law,
F = ke................. Equation 1
Where F = Force required to stretch the spring, k = spring constant, e = extension.
make k the subject of the equation
k = F/e.............. Equation 2
Given: F = 10 lb = (10×4.45) N = 44.5 N, e = 4 in = (4×0.254) = 1.016 m.
Substitute into equation 2
k = 44.5/1.016
k = 43.799 N/m
Work done in stretching the 9 in beyond its natural length
W = 1/2ke²................. Equation 3
Given: e = 9 in = (9×0.254) = 2.286 m, k = 43.799 N/m
Substitute into equation 3
W = 1/2×43.799×2.286²
W = 114.44 J
The formula for working out speed is distance ÷ time.
55 km ÷ 2 hours = 27.5 km/h (average speed for first part of journey)
52km ÷ 5 hours = 10.4 km/h (average speed for second part of journey)
(27.5 + 10.4) ÷ 2 = 18.95 km/h (average speed throughout the journey)
Answer:
the answer is calcium....
Answer
given,
firm is producing = 2,475 units
output by hiring 50 workers W = $20 per hour
25 units of capital R = $10 per hour
marginal product of labor = 40
marginal product of capital = 25
Firm is not minimizing the cost because the firm use more capital and less labor.
