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:
13.4cm
Explanation:
According to Rayleigh’s criterion the angular resolution to distinguish two objects is given by:
θ = 50.0*10^-7 rad
λ: wavelength of the light = 550nm
b = diameter of the objective
By doing b the subject of the formula and replacing the values of the angle and wavelength you obtain:
hence, the smallest diameter objective lens is 13.4cm
Answer:
It grows
Explanation:
The blacks holes will absorb
Me hoizontally stretching me like a noodle by the spaghtification process,thus growing bigger.
C is the answer to the question
Hooke's Law states that the extension is directly proportional to the force applied so:
F/x = constant
F₁/x₁ = F₂/x₂
2 / 0.02 = 1600 / x₂
x₂ = 16 m
Elastic work = 1/2 Fx
= 1/2 * 1600 * 16
= 12.8 kJ
