Answer:
Explanation:
Total length of the wire is 29 m.
Let the length of one piece is d and of another piece is 29 - d.
Let d is used to make a square.
And 29 - d is used to make an equilateral triangle.
(a)
Area of square = d²
Area of equilateral triangle = √3(29 - d)²/4
Total area,
Differentiate both sides with respect to d.
For maxima and minima, dA/dt = 0
d = 8.76 m
Differentiate again we get the
(a) So, the area is maximum when the side of square is 29 m
(b) so, the area is minimum when the side of square is 8.76 m
Answer:
D
Explanation:
P=Work/Time
The rate at which work is done matches that.
Explanation:
Work done is given by the product of force and displacement.
Case 1,
1. A boy lifts a 2-newton box 0.8 meters.
W = 2 N × 0.8 m = 1.6 J
2. A boy lifts a 5-newton box 0.8 meters.
W = 5 N × 0.8 m = 4 J
3. A boy lifts a 8-newton box 0.2 meters.
W = 8 N × 0.2 m = 1.6 J
4. A boy lifts a 10-newton box 0.2 meters.
W = 10 N × 0.2 m = 2 J
Out of the four options, in option (2) ''A boy lifts a 5-newton box 0.8 meters'', the work done is 4 J. Hence, the greatest work done is 4 J.