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
When angle between them is zero
Answer:
The greatest acceleration the man can give the airplane is 0.0059 m/s².
Explanation:
Given that,
Mass of man = 85 kg
Mass of airplane = 109000 kg
Distance = 9.08
Coefficient of static friction = 0.77
We need to calculate the greatest friction force
Using formula of friction

Where, m = mass of man
g = acceleration due to gravity
Put the value into the formula


We need to calculate the acceleration
Using formula of newton's second law


Put the value into the formula


Hence, The greatest acceleration the man can give the airplane is 0.0059 m/s².
C. High-visibility clothing is important to wear in areas with moving vehicles
Answer:
Explanation:
Given
average speed of train
Maximum acceleration=0.05g
Now centripetal acceleration is


r=7346.93 m
(b)Radius of curvature=900 m
therefore 


