The total work done is 5980 Joules and the power expended is 57 Watts.
<h3>What is work done?</h3>
The work done is the work done in the gravitational field as the bucket is raised up Thus work required to remove the bucket Wb;
Wb = 13.9 kg * 25.9 m * 9.8 m/s^2 = 3530 Joules
Height of the center of mass of chain = 25.9 / 2 = 12.95 m
Work done by the chain Wc;
Wc = 12.95 * 19.3 * 9.8 = 2450 Joules
Total work = 3530 + 2450 = 5980 Joules
Power expended = W / t = 5980 J / 105 sec = 57 J/s = 57 Watts
Learn more about work done:brainly.com/question/13662169
#SPJ1
Actual Mechanical Advantage(AMA) = Weight / Force
Here, Weight = 764 N
Force = 255 N
Substitute the values in to the expression,
AMA = 764 / 255
AMA = 2.99
After rounding-off to the nearest tenth value, it would be 3
Finally, option C would be your answer.
Hope this helps!
Answer:
a)1815Joules b) 185Joules
Explanation:
Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;
F = ke where;
F is the applied force
k is the elastic constant
e is the extension of the material
From the formula, k = F/e
F1/e1 = F2/e2
If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;
k = 60/0.5
k = 120N/m
a) To get the work done in stretching the spring 5.5m from its position,
Work done by the spring = 1/2ke²
Given k = 120N/m, e = 5.5m
Work done = 1/2×120×5.5²
Work done = 60× 5.5²
Work done = 1815Joules
b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;
Work done = 1/2ke²
Work done =1/2× 120×1.5²
Works done = 60×1.5²
Work done = 135Joules
X=1/2 at^2
3.1=1/2 a *0.64
a=9.68
v=at
v=0.8*9.6875=7.75
Answer:
brand name, or company
Explanation:
a brand name is the kind of product they are selling. a company is the group of peaple that wok together.
