Answer:
Work done in stretching the spring = 7.56 lb-ft.
Step-by-step explanation:
Normal length of the spring = 8 in or 
ft
= 
ft
If the spring has been stretched to 11 inch then the stretched length of the spring is = 11 in
= 
ft
Force applied to stretch the spring = 12 lb
By Hook's law,
F = kx [where k is the spring constant and x = length by which the spring is stretched]
12 = k()
k = 18
Work done (W) to stretch the spring by ft will be
W = 
= 
= ![18[\frac{x^{2}}{2}]^{\frac{11}{12}}_0](https://tex.z-dn.net/?f=18%5B%5Cfrac%7Bx%5E%7B2%7D%7D%7B2%7D%5D%5E%7B%5Cfrac%7B11%7D%7B12%7D%7D_0)
= 9()²
= 7.56 lb-ft
Answer: 500m
Explanation:
If the scale of the drawing is 1:500 then the real playground should be 500:1 which is 500m
The rule of exponential multiplication is (a^x)*(a^y)=a^(x+y)
using this with your expression, we can do
(10^1)*(10^3)*(10^2)=10^(1+3+2)=10^6
Final answer:
10^6 or 1 million
Hope I helped :)
Wa=0.5Wb, Ha=20, Hb is unkown
the area of rectangle a is : Wa*20
the area of rectangle b is : Wb*Hb
Rectangle a has 7 times the area of rectangle b: Wa*20=7*Wb*Hb
Replace Wa with 0.5Wb: 0.5Wb*20=7Wb*Hb => Hb=10/7
Answer:
the anwers is 5
Step-by-step explanation:
