He goes 12km north 6km east 12km south 3km west what is the distance and displacement
1 answer:
Distance is 33 km, displacement is 3 km east
Explanation:
- Distance is a scalar quantity equal to the total length of the path covered by an object during its motion, regardless of its direction.
- Displacement is a vector connecting the starting position of motion of an object to its final position. It is a vector quantity, therefore it has both magnitude and direction.
In this problem, the motion of the person is:
12 km north
6 km east
12 km south
3 km west
Therefore, the distance is:
The displacement instead can be calculated by comparing the final position with the initial position. We notice that:
- the person goes 12 km north and 12 km south, so he didn't move in the north-direction
- The person goes 6 km east and 3 km west, so his final position is 3 km east
Therefore, the displacement is 3 km east.
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Answer:
The value of the centripetal forces are same.
Explanation:
Given:
The masses of the cars are same. The radii of the banked paths are same. The weight of an object on the moon is about one sixth of its weight on earth.
The expression for centripetal force is given by,
where, is the mass of the object,
is the velocity of the object and
is the radius of the path.
The value of the centripetal force depends on the mass of the object, not on its weight.
As both on moon and earth the velocity of the cars and the radii of the paths are same, so the centripetal forces are the same.
Answer:
Fnet = 14515.2 Newton
Explanation:
Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.
Mathematically, net force is given by the formula;
Where;
- Fnet is the net force.
- Fapp is the applied force.
- Fg is the force due to gravitation.
Given the following data;
Mass = 1,152 kg
Initial velocity, u = 3m/s
Final velocity, v = 17m/s
Time, t = 5 seconds
To find the magnitude of the net force;
First of all, we would determine the acceleration of the car.
Acceleration = (v-u)/t
Acceleration = (17 - 3)/5
Acceleration = 14/5
Acceleration = 2.8m/s
To find the applied force;
Fapp = mass * acceleration
Fapp = 1,152 * 2.8
Fapp = 3225.6 N
Next, we would find the force exerted on the car due to gravity.
Fg = mass * acceleration due to gravity
We know that acceleration due to gravity is equal to 9.8m/s²
Fg = 1152 * 9.8
Fg = 11289.6N
Substituting the values into the equation, we have;
Fnet = 3225.6 + 11289.6
Fnet = 14515.2 Newton