From 1st EQ. of uniform motion
1)20.9-3.44/6=t=2.91 sec
2)28= 0+4.22a
a=6.66m/sec^2
3) Instantaneous acceleration
Answer:
F_total = 10000 N
Explanation:
For this exercise we use that the force is a vector and the best way to do the sum is that since the two tugs pull the boat with the same intensity and angle, the sum of the component of the force perpendicular to the movement becomes zero.
The components parallel to the movement of the tugs is
∑ F = 2 Fcos θ
let's calculate
F_total = 2 10000 cos 60
F_total = 10000 N
The block gains gravitational potential energy as it gets higher,
and that's the energy that the motor has to supply.
The amount of potential energy gained by the block is
(mass) x (gravity) x (height)
= (2.0 kg) x (9.8 m/s²) x (15 m)
= (2 x 9.8 x 15) kg-m²/s² = 294 joules.
Power = (change in energy) / (time for the change)
= (294 joules) / (6 sec)
= 49 joules/second = 49 watts .
Initial speed(u)=0m/s
Final speed(v)= 27m/s
Time(t)=7.6s
Use the equation of motion: v = u + at
27 = 0 + a(7.6)
27/7.6 = a
a = 3.55 m/s^2 (3 s.f)
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am hoping that this answer has satisfied your query and it will be able to help
you in your endeavor, and if you would like, feel free to ask another question.
