Given Data: Diameter 'd' = 30 cm = 0.3 m Lifting Weight 'W' = mg = 2000*9.81 N = 19,620 N Calculations: Area of the lift 'A' = <span>pi\over4*d^2=pi\over4*0.3^2=0.07 m^2
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Answer:
29.16 J
Explanation:
From Hook's law,
W = 1/2(ke²)..................... Equation 1
Where W = work done, k = Spring constant, e = extension.
Given: W = 9 J, e = 0.5 m.
Substitute into equation 1
9 = 1/2(k×0.5²)
Solve for k
k = 18/0.5²
k = 72 N/m.
The work done required to stretch the spring by additional 0.4 m is
W = 1/2(72)(0.4+0.5)²
W = 36(0.9²)
W = 29.16 J.
Slide with her left foot. hope this is helpful
The power is 833.3 W
Explanation:
First of all, we need to calculate the work done in lifting the barbell, which is equal to the change in gravitational potential energy of the barbell:
where
mg = 1250 N is the weight of the barbell
h = 2 m is the change in height
Substituting,
Now we can calculate the power, which is equal to the work done per unit time:
where
W = 2500 J is the work done
t = 3 s is the time taken
Substituting,
Learn more about power:
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A good way for me to remember things is to study it, and to write it down! Say you want the formula for speed, I would write the formula multiple times on a piece of paper!
Here's a video that I haven't actually watched, I just looked it up! It might help you out though: <span>https://www.youtube.com/watch?v=-Wqrw4G79Kc</span>
