1. the pilot of a private plane flies 20km in a direction 60 degrees north of east,then 30km straight east,then 10km straight north.how far and in what direction is the plane from the starting point.
2. two forces acts on a point object as follows 100N at 170 degrees and 100N at 50 degrees.find their resultant.
3. A car is traveling eastwards at a speed of 40m/s.But a 3m/s wind is blowing southward.What are the direction and speed of the car relative to its original direction.
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1. To solve this problem, we can break the distances and directions into components and use vector addition to find the resultant vector.
First, let's break down the given distances into their corresponding x and y components:
20km at 60 degrees north of east:
The x component is 20km * cos(60) = 10 km
The y component is 20km * sin(60) = 17.32 km (approx)
30km straight east:
30 km east, so the x component is 30 km and the y component is 0.
10km straight north:
10 km north, so the x component is 0 and the y component is 10 km.
Now, let's add up the x and y components to find the resultant:
Summing up the x components, we have: 10 km + 30 km + 0 km = 40 km east
Summing up the y components, we have: 17.32 km + 0 km + 10 km = 27.32 km north
Now we can calculate the magnitude of the resultant:
Resultant distance = sqrt((40 km)^2 + (27.32 km)^2) ≈ 48.15 km
To find the direction of the plane from the starting point, we can use trigonometry:
Resultant angle = arctan(27.32 km / 40 km) ≈ 34.29 degrees north of east
Therefore, the plane is approximately 48.15 km away from its starting point, in the direction 34.29 degrees north of east.
2. To find the resultant force from two forces acting on an object, we can use vector addition.
First, let's break down the given forces into their corresponding x and y components:
100N at 170 degrees:
The x component is 100N * cos(170) = -98N (approx)
The y component is 100N * sin(170) = 34N (approx)
100N at 50 degrees:
The x component is 100N * cos(50) = 64.28N (approx)
The y component is 100N * sin(50) = 76.6N (approx)
Now, let's add up the x and y components to find the resultant:
Summing up the x components, we have: -98N + 64.28N = -33.72N (approx)
Summing up the y components, we have: 34N + 76.6N = 110.6N (approx)
Now we can calculate the magnitude of the resultant force:
Resultant magnitude = sqrt((-33.72N)^2 + (110.6N)^2) ≈ 115.55N
To find the direction of the resultant force, we can use trigonometry:
Resultant angle = arctan(110.6N / -33.72N) ≈ -73.58 degrees
Therefore, the resultant force acting on the object is approximately 115.55N at an angle of -73.58 degrees (measured counterclockwise from the positive x-axis).
3. To determine the direction and speed of the car relative to its original direction, we can use vector addition.
The car is traveling eastward at a speed of 40 m/s, so its velocity can be represented as a vector of magnitude 40 m/s in the positive x-direction.
The wind is blowing southward at a speed of 3 m/s, so its velocity can be represented as a vector of magnitude 3 m/s in the negative y-direction (opposite to the positive y-axis).
To find the resultant velocity, we need to add the velocities of the car and the wind.
Since the wind acts perpendicular to the car's original velocity, we can simply add the x-components and y-components separately.
Summing up the x-components, we have: 40 m/s + 0 m/s = 40 m/s
Summing up the y-components, we have: 0 m/s + (-3 m/s) = -3 m/s
The magnitude of the resultant velocity is given by:
Resultant speed = sqrt((40 m/s)^2 + (-3 m/s)^2) ≈ 40.03 m/s
To determine the direction of the resultant velocity (relative to the car's original direction), we can use trigonometry:
Resultant angle = arctan((-3 m/s) / 40 m/s) ≈ -4.27 degrees
Therefore, the car's resultant velocity (relative to its original direction) is approximately 40.03 m/s at an angle of -4.27 degrees (measured counterclockwise from its original direction).