The work done to pull the object 7.0 m is the total area under the graph from 0.0 m to 7.0 m, determined as 245 J.
<h3>Work done by the applied force</h3>
The area under force versus displacement graph is work done.
The total work done by pulling the object 7 m, can be grouped into two areas;
- First area, A1 = area of triangle from 0 m to 2.0 m
- Second area, A2 = area of trapezium, from 2.0 m to 7.0 m
A1 = ¹/₂ bh
A1 = ¹/₂ x (2) x (20)
A1 = 20 J
A2 = ¹/₂(large base + small base) x height
A2 = ¹/₂[(7 - 2) + (7-3)] x 50
A2 = ¹/₂(5 + 4) x 50
A2 = 225 J
<h3>Total work done </h3>
W = A1 + A2
W = 20 J + 225 J
W = 245 J
Learn more about work done here: brainly.com/question/8119756
Answer: the contents of this container weighs 4905 kg.m/s²
Explanation:
Given that;
volume of a container V = 0.5 m³
we know that standard gravitational acceleration g = 9.81 m/s²
specific volume of liquid filled in the container v = 0.001 m³/kg
now we express the equation for weight of the container.
W = mg
W = (pV)g
W = Vg / ν
so we substitute
W = (0.5 m³)(9.81 m/s ) / 0.001 m³/kg
W = 4.905 / 0.001
W = 4905 kg.m/s²
Therefore, the contents of this container weighs 4905 kg.m/s²
Answer:
The answer is D- Coal is set aside to burn as heating fuel
Explanation:
Allocation decisions set aside resources for a specific purpose.
Answer:
5 moles of O2are required, you can see it in your equation.
64 miles/hour
Therefore 1/64 hours/mile
68 miles * 1/64 hours/mile (notice how miles cancels out)
Therefore the answer is 68/64 hours = 1.0625 hours = 1 hour 3min and 45sec.
