Answer:corrosion (i believe)
Explanation:
Answer:
0.56 m/s
Explanation:
The speed of the head at the end of the interval in each case is the area under the acceleration curve. Then the difference in speeds is the difference in areas.
We can find the area geometrically, using formulas for the area of a triangle and of a trapezoid.
A = 1/2bh . . . . area of a triangle
A = 1/2(b1 +b2)h . . . . area of a trapezoid
If h(t) is the acceleration at time t for a helmeted head, the area under that curve will be (in units of mm/s) ...
Vh = 1/2(h(3)·3) +1/2(h(3) +h(4))·1 +1/2(h(4) +h(6))·2 +1/2(h(6))·1
Vh = 1/2(4h(3) +3h(4) +3h(6)) = 1/2(4·40 +3·40 +3·80) = 260 . . . mm/s
If b(t) is the acceleration for a bare head, the area under that curve in the same units is ...
Vb = 1/2(b(2)·2 +1/2(b(2) +b(4))·2 +1/2(b(4) +b(6))·2 +1/2(b(6)·1)
Vb = 1/2(4b(2) +4b(4) +3b(6)) = 1/2(4·120 +4·140 +3·200) = 820 . . . mm/s
Then the difference in speed between the bare head and the helmeted head is ... (0.820 -0.260) m/s = 0.560 m/s
To show stuff that we cant see in very well
Answer:
Total work done = 15,306.25 lb.ft
Explanation:
First of all, we know that;
Work done = Force x Distance.
Thus,
Work done to pull bucket (Wb) = 500 x 30 = 15,000 lb.ft
Now, work done by pulling the rope is given as a function of the length of the rope. Thus;
Wr =(x,x=0∫) F(x) dx
=(35,0∫) 0.5x dx
= 0.5[x²/2](35,0)
Thus,Wr = 0.5[(35²/2) - (0²/2)] = 0.5[612.5] = 306.25 lb.ft
Total work done will be the sum of that done to pull the bucket and that done to pull the rope.
Thus, Wt = Wb + Wr
Wt = 15000 + 306.25 = 15,306.25 lb.ft