Answer:
The work done in stretching the spring is 8 lb.ft
Step-by-step explanation:
Given;
Applied force, F = 192 lb
extension of the spring, x = 3 ft
Determine the spring constant from the applied force and extension;
When the spring is stretched 6 inches beyond its natural length, the work done is calculated as follows;
x = 6 inches = 0.5 ft
Therefore, the work done in stretching the spring is 8 lb.ft
Answer:
<h2>250%</h2>
Step-by-step explanation:
So first lets divide!
10÷4=2.5
Now, 2.5 x 100 = 250!
Hope this helped! Have a great day!
H(t) = -16t² + 60t + 95
g(t) = 20 + 38.7t
h(1) = -16(1²) + 60(1) + 95 = -16 + 60 + 95 = -16 + 155 = 139
h(2) = -16(2²) + 60(2) + 95 = -16(4) + 120 + 95 = -64 + 215 = 151
h(3) = -16(3²) + 60(3) + 95 = -16(9) + 180 + 95 = -144 + 275 = 131
h(4) = -16(4²) + 60(4) + 95 = -16(16) + 240 + 95 = -256 + 335 = 79
g(1) = 20 + 38.7(1) = 20 + 38.7 = 58.7
g(2) = 20 + 38.7(2) = 20 + 77.4 = 97.4
g(3) = 20 + 38.7(3) = 20 + 116.1 = 136.1
g(4) = 20 + 38.7(4) = 20 + 154.8 = 174.8
Between 2 and 3 seconds.
The range of the 1st object is 151 to 131.
The range of the 2nd object is 97.4 to 136.1
h(t) = g(t) ⇒ 131 = 131
It means that the point where the 2 objects are equal is the point where the 1st object is falling down while the 2nd object is still going up.
Answer:
60 degrees.
Step-by-step explanation:
We use trigonometry of a right angled triangle:
Adjacent side = 20 ft, Hypotenuse = 40 ft.
cos x = adjacent / hypotenuse = 20/40 = 1/2
x = 60 degrees.
