Answer:
0.29 m/s due west.
Explanation:
According to newton's second law,
Net force acting on an object = mass×acceleration
From the question,
F+F₁+F₂ = ma................ Equation 1
Where F = The force generated from the engine, F₁ = Force exerted by the wind, F₂ = Force exerted due to the water, m = mass of the boat, a = acceleration of the boat.
Given: F = 4080 N , F₁ = -680 N(east), F₂ = -1160 N(east). m = 7660 kg
substitute into equation 1
4080-680-1160 = 7660(a)
2240 = 7660a
Therefore,
a = 2440/7660
a = 0.29 m/s due west.
Answer:
Answer: <u>Height</u><u> </u><u>is</u><u> </u><u>0</u><u>.</u><u>2</u><u>0</u><u>4</u><u> </u><u>m</u>
Explanation:
At the highest point, it is called the maximum height.
• From third newton's equation of motion:

• At maximum height, v is zero
• u is initial speed
• g is -9.8 m/s²
• s is the height

Missing figure: http://d2vlcm61l7u1fs.cloudfront.net/media/f5d/f5d9d0bc-e05f-4cd8-9277-da7cdda3aebf/phpJK1JgJ.png
Solution:
We need to find the magnitude of the resultant on both x- and y-axis.
x-axis) The resultant on the x-axis is

in the positive direction.
y-axis) The resultant on the y-axis is

in the positive direction.
Both Fx and Fy are positive, so the resultant is in the first quadrant. We can find the angle and so the direction using

from which we find
Answer:
Time of flight A is greatest
Explanation:
Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.
So
H = u₁² sin²θ₁ /2g
H = u₂² sin²θ₂ /2g
H = u₃² sin²θ₃ /2g
On the basis of these equation we can write
u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃
For maximum range we can write
D = u₁² sin2θ₁ /g
1.5 D = u₂² sin2θ₂ / g
2 D =u₃² sin2θ₃ / g
1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁
1.5 = u₂ cosθ₂ /u₁ cosθ₁ ( since , u₁ sinθ₁ =u₂ sinθ₂ )
u₂ cosθ₂ >u₁ cosθ₁
u₂ sinθ₂ < u₁ sinθ₁
2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g
Time of flight B < Time of flight A
Similarly we can prove
Time of flight C < Time of flight B
Hence Time of flight A is greatest .