a^2 = b^2 + c^2
a^2 = (3000)^2 + (6000)^2 - 2(3000)(6000)
a^2 = 45000000 - 36000000
a = 4420.877N
F = ma 4420.877N / 2000kg = 2.2m/s^2
a = F/m
a^2 = (3000)^2 + (6000)^2 - 2(3000)(6000)
a^2 = 45000000 - 36000000
a = 4420.877N
F = ma 4420.877N / 2000kg = 2.2m/s^2
a = F/m
First, let's break down the given forces into their x and y components.
1. Ocean tide force (3000N):
- This force is acting eastward, so its x-component is 3000N.
- Since this force is not directed north or south, its y-component is 0N.
2. Wind force (6000N) at 45 degrees north of west:
- To determine the x and y components, we can use trigonometry.
- The magnitude of the x-component is given by:
x-component = 6000N * cos(45°)
x-component = 6000N * 0.7071
x-component = 4242.64N (rounded to four decimal places)
- The magnitude of the y-component is given by:
y-component = 6000N * sin(45°)
y-component = 6000N * 0.7071
y-component = 4242.64N (rounded to four decimal places)
Now, let's find the net force by adding the x and y components separately.
Net x-component = Ocean tide x-component + Wind x-component
= 3000N + 4242.64N
= 7242.64N (rounded to two decimal places)
Net y-component = Ocean tide y-component + Wind y-component
= 0N + 4242.64N
= 4242.64N (rounded to two decimal places)
Next, we can find the magnitude of the resultant force using the Pythagorean theorem:
Resultant force (F) = sqrt((Net x-component)^2 + (Net y-component)^2)
= sqrt((7242.64N)^2 + (4242.64N)^2)
= sqrt(52608280.52N^2)
= 7254.61N (rounded to two decimal places)
Finally, we can calculate the magnitude of the resultant acceleration using Newton's second law:
F = ma
7254.61N = (200kg) * a
a = 7254.61N / 200kg
= 36.27 m/s² (rounded to two decimal places)
Therefore, the magnitude of the resultant acceleration is approximately 36.27 m/s².
In this case, the sailboat experiences two forces: the eastward force by the ocean tide (3000N) and the wind force against its sails (6000N) directed towards the northwest (45 degrees north of west). To determine the resultant force, we need to find the vector sum of these two forces.
First, we can break down the wind force into horizontal and vertical components. Since the wind force is directed towards the northwest at a 45-degree angle north of west, we can use trigonometry to find its components.
The horizontal component (F_horizontal) can be calculated using the cosine function:
F_horizontal = F_wind * cos(45 degrees)
The vertical component (F_vertical) can be calculated using the sine function:
F_vertical = F_wind * sin(45 degrees)
Given that the wind force (F_wind) is 6000N, we can substitute this value into the above equations to calculate the horizontal and vertical components.
F_horizontal = 6000N * cos(45 degrees)
F_vertical = 6000N * sin(45 degrees)
Now we can determine the resultant forces in the horizontal (Fx) and vertical (Fy) directions by adding the forces acting in those directions:
Fx = F_horizontal + 3000N (since the eastward force only acts in the horizontal direction)
Fy = F_vertical
Finally, we can find the magnitude of the resultant acceleration (a) using Newton's second law:
Resultant force (F_resultant) = sqrt(Fx^2 + Fy^2)
a = F_resultant / mass
Substituting the values we calculated, we can find the magnitude of the resultant acceleration.