Answer:
The distance it has traveled is 3,050 m and the magnitude of its displacement is 650 m north.
Explanation:
Distance refers to the length between any two points in space, while displacement refers to the distance from a start position to an end position regardless of the path.
In other words, distance refers to how much space an object travels during its movement; is the quantity moved. It is also said to be the sum of the distances traveled. The distance traveled by a mobile is the length of its trajectory and it is a scalar quantity. In this case, the distance is calculated as:
1850 m + 1200 m= 3,050 m
Displacement refers to the distance and direction of the final position from the initial position of an object. The displacement effected is a vector quantity. The vector representing the displacement has its origin in the initial position, its end in the final position, and its module is the distance in a straight line between the initial and final positions. That is, when expressing the displacement it is done in terms of the magnitude with its respective unit of measurement and the direction because the displacement is a vector type quantity. Mathematically, the displacement (Δd) is calculated as:
Δd= df - di
where df is the final position and di is the initial position of the object.
In this case, the displacement is calculated as:
1850 m - 1200 m= 650 m
Since the distance to the north is greater, the direction of travel will be to the north.
<u><em>The distance it has traveled is 3,050 m and the magnitude of its displacement is 650 m north.</em></u>
Answer:
a) 0.138J
b) 3.58m/S
c) (1.52J)(I)
Explanation:
a) to find the increase in the translational kinetic energy you can use the relation
where Wp is the work done by the person and Wg is the work done by the gravitational force
By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:
the change in the translational kinetic energy is 0.138J
b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:
where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:
the new speed is 3.58m/s
c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy
hence, the change in Er is about 1.52J times the initial rotational energy
Medical movement for disabilities people
So the problem are asking to find the value of G base on the formula of the said equation of the magnitude of gravitational attraction on either body. Base on that, the possible answer or the derived formula of the said function is G = Fr^2/m1m2. I hope you are satisfied with my answer and feel free to ask for more
