Answer:
not sure. I'll try answering this later
Explanation:
I'm not sure. I'll try answering this later .
a. We can calculate the amount of work by calculating the area under the graph.
first area (rectangular): 2.5 x 6 = 15
second area(trapezoid): 1/2 x (6+10) x 2.5 =20
total work done: 35 J
b. the force was first applied = 6 N
F = m.a
a = 6 : 3 = 2 m/s²
vf²=vi²+2as
vf²=6²+2.2.5
vf²=56
vf=7.5 m/s
Fortunately, 'force' is a vector. So if you know the strength and direction
of each force, you can easily addum up and find the 'resultant' (net) force.
When we talk in vectors, one newton forward is the negative of
one newton backward. Hold that thought, while I slog through
the complete solution of the problem.
(100 N forward) plus (50 N backward)
= (100 N forward) minus (50 N forward)
= 50 N forward .
That's it.
Is there any part of the solution that's not clear ?
Answer:
everything is perished isn't it?
hope it helps.
The answer for this is actually b. 79 m/s
I submitted it to odyssey-ware and i got the question correct
