1. Two people are pushing a disabled car. One exerts a force of 200 N East, the other a force of 250 N East. What is the net force exerted on the car?
2. Two hockey players strike a puck simultaneously from opposite sides. Red #3 hits it with 40 N of force to the right, while Blue #5 hits it with 63 N. What is the net force on the puck?
3. An airplane heads due north at 100 m/s through a 30 m/s cross wind blowing from east to west. Determine the resultant velocity of the airplane.
add the vectors with direction: ie, 10N+20S+43E=(10-20)N+43E. If the result is in two directions, since they are at 90 degrees, you will have a resultant at an angle. You can do the angle graphically, or with trig
angle(from N) = arctan( E/N) where E and N are those components.
Magnitude=sqrt(N^2 + E^2) or Magnitude=E/sinAngle
as a general rule, I recommend using geographic coordiantes, measuring angles from 000 (N), clockwise (090 is E); 180 is S, 270 is W. It greatly helps you find the angles.
1. To determine the net force exerted on the car, we need to add the forces together. In this case, both forces are acting in the same direction (East), so we can simply add the magnitudes of the forces.
Net force = 200 N + 250 N = 450 N East
Therefore, the net force exerted on the car is 450 N East.
2. Similar to the previous question, we need to add the forces together. However, in this case, the forces are acting in opposite directions (right and left). To find the net force, we need to subtract one force from the other.
Net force = 63 N - 40 N = 23 N to the right
Therefore, the net force on the puck is 23 N to the right.
3. To find the resultant velocity of the airplane, we need to consider both the northward velocity and the crosswind effect.
Using the Pythagorean theorem, we can find the magnitude of the resultant velocity:
Resultant velocity = √(Northward velocity^2 + Crosswind velocity^2)
= √(100 m/s)^2 + (30 m/s)^2)
≈ √(10000 m^2/s^2 + 900 m^2/s^2)
≈ √10900 m^2/s^2
≈ 104.4 m/s
Therefore, the resultant velocity of the airplane is approximately 104.4 m/s.
1. To find the net force exerted on the car, we need to add the individual forces together. The net force is the sum of all the forces acting on an object.
In this case, one person exerts a force of 200 N East, and the other person exerts a force of 250 N East. Since both forces are acting in the same direction, we can simply add them to find the net force.
Net force = 200 N + 250 N = 450 N East
Therefore, the net force exerted on the car is 450 N East.
2. Similar to the first question, we need to add the individual forces to find the net force on the puck.
Red #3 strikes the puck with a force of 40 N to the right, and Blue #5 strikes it with a force of 63 N. Since the forces are acting in opposite directions, we need to consider their signs when adding them.
Net force = 40 N (to the right) - 63 N = -23 N
Therefore, the net force on the puck is 23 N to the left (since -23 N is a force in the opposite direction).
3. To find the resultant velocity of the airplane, we need to consider both its velocity and the velocity of the crosswind.
The airplane is heading due north at 100 m/s, and the crosswind is blowing from east to west at a velocity of 30 m/s. We can break down the velocities into their respective components using vector addition.
The northward velocity of the airplane remains the same since there is no force acting in that direction.
The eastward velocity of the crosswind is 30 m/s, and the airplane is not affected by any opposing force in that direction.
So, the resultant velocity is the vector sum of the northward velocity (100 m/s) and the eastward velocity (30 m/s).
By using the Pythagorean theorem, we can calculate the magnitude of the resultant velocity:
Resultant velocity = sqrt((100 m/s)^2 + (30 m/s)^2) ≈ 104 m/s
The direction of the resultant velocity can be found using trigonometry. The tangent of the angle formed between the northward and eastward directions is given by the ratio of the two velocities:
tanθ = 30 m/s / 100 m/s
θ ≈ 16.7 degrees
Therefore, the resultant velocity of the airplane is approximately 104 m/s at an angle of 16.7 degrees north of east.