Mathematics, 29.06.2020 02:01 sarahsompayrac

Vector Application Handout (First one to answer correctly gets brainest)

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Answer from: devinmoore4664

1. 50·√2·i - 50·√2·j

2. The magnitude is 220.49 pounds, the direction 16.7°

3. 197.9 N

4. The resultant speed is 778.31 km/hr

The true course of the plane is 27.72°

5. The direction of the boat is 354.315°

The speed of the boat is 34.19 knot

Step-by-step explanation:

1. Given that the airplane is descending at 100 mph at an angle 45° below the horizontal, we have;

Horizontal component = 100 × cos(45°) = 100 × (√2)/2 = 50·√2 mph

The vertical component = 100 × sin(-45°) = 100 × (-√2)/2 = -50·√2 mph

Which gives;

50·√2·i - 50·√2·j

2. The given forces are;

F₁ = 250 pounds at 25° with the horizontal

F₁ₓ = 250 × cos(25°) = 226.58 pounds

$F_{1y}$ = 250 × sin(25°) = 105.655 pounds

F₂ = 45 pounds at 250° with the horizontal

F₂ₓ = 45 × cos(250°) = -15.39 pounds

$F_{2y}$ = 45 × sin(250°) = -42.29 pounds

The magnitude of the resultant force F = √((F₁ₓ - F₂ₓ)² + ( $F_{1y}$ - $F_{2y}$ )²)

F = √((226.58 + (-15.39))² + (105.655 + (-42.29))²)

F = √(211.19)² + (63.365)²) = 220.49 pounds

The direction of the resultant force θ = $tan^{-1} \left (\dfrac{F_y}{F_x} \right) = tan^{-1} \left (\dfrac{63.365}{211.19} \right) = 16.7^ {\circ}$

3. The horizontal force = 120 × cos(60°) + 180 × cos(40°) = 197.9 N

4. Direction of airplane = 30° North of East. we have;

Speed of airplane, $S_{A}$ = 724 km/hr

The x component of the speed of the airplane, $S_{Ax}$ = 724 × cos(30°) = 627 km/hr

The y component of the speed of the airplane, $S_{Ay}$ = 724 × sin(30°) = 362 km/hr

The wind speed, S_w = $S_{wx}$ = 62 km/hr

The resultant speed in the x direction, Sₓ = $S_{Ax}$ + $S_{wx}$ = 627 + 62 = 689 km/hr

The resultant speed in the y direction, $S_{y}$ = 362 km/hr

The magnitude of the resultant speed, S = √(Sₓ² + $S_{y}$ ²) = √(689² + 362²) = 778.31 km/hr

The direction of the airplane θ = $tan^{-1} \left (\dfrac{S_y}{S_x} \right) = tan^{-1} \left (\dfrac{362}{689} \right) = 27.72^ {\circ}$

5. The x, $S_{yx}$ and y, $S_{yy}$ components of the speed of the yacht are;

$S_{yy}$ = 20×sin(185°) = -1.743 knots

$S_{yx}$ = 20×cos(185°) = -19.924 knots

The x, $S_{wx}$ and y, $S_{wy}$ components of the speed of the yacht are;

$S_{wx}$ = 15×sin(290°) = -14.1 knots

$S_{wy}$ = 15×cos(290°) = 5.13 knots

The resultant x, $S_{x}$ , and y, $S_{y}$ are found as follows;

$S_{x}$ = $S_{wx}$ + $S_{yx}$ = -19.924 + -14.1 = -34.024 knots

$S_{y}$ = $S_{yy}$ + $S_{wy}$ = -1.743 + 5.13 = 3.387

The resultant speed is given by the relation;

S =√( $S_{x}$ ² + $S_{y}$ ²) = √((-34.024)^2 + 3.387^2) = 34.19 knot

The direction of the boat, θ = $tan^{-1} \left (\dfrac{S_y}{S_x} \right) = tan^{-1} \left (\dfrac{3.387}{-34.024} \right) = -5.685^ {\circ}$

The direction of the boat = -5.685 = 354.315°

The speed of the boat = 34.19 knot.

Answer from: Quest

the first one, all those points on the line just make a strait line.

Which table represents a linear function?

Answer from: Quest

answer: 0,6

step-by-step explanation:

this would make it the y intercept.

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