Answer:
Light has the unique property that it can be described in physics as both a wave and as a stream of particles called photons
Explanation:
Answer:A force can be described as a push or a pull. Pushes and pulls can be seen to act on objects when they begin to move, speed up, slow down or change direction.
The image gallery on this page shows children at play using push and pull actions to move objects. Children use push and pull all the time while at play. While playing, students learn to manipulate objects and materials and make observations about their actions.
Teachers may use this teaching resource over a number of lessons to explore and develop their students' understanding that a push or a pull affects how an object moves or changes shape.
Explanation:
Answer:
a)
The magnitude of the attractive force is
b)
The magnitude of the repulsive force is
c)
The magnitude of the net force is
Explanation:
The explanation is shown on the first and second uploaded image
Answer:
Positive
Explanation:
Work is defined as the product of displacement of an object produced by the applied force and the component of force along that direction. Let us consider a situation in which the angle between the applied force and the displacement produced by it is at an angle θ. Let the magnitude of displacement be s and that of the applied force be F. Now, we have to find the component of force along the direction of the displacement. For that we have to resolve the force in to two components- one along the direction of displacement and other perpendicular to it. Since the angle between force and displacement is θ, the component of force along the direction of displacement will be Fcosθ and that perpendicular to it will be Fsinθ. Thus, b the definition of work, work done = Fcosθ x s = Fscosθ.
Now, coming to our question, the force here is gravitational force of attraction which is along downward direction. It is given in the question itself that the stone is falling down. Since the displacement and the applied force is along the same direction, angle between them, θ = 0.
Thus, work done = Fscosθ = Fscos0 = Fs (since cos0 =1)
Fs > 0
Thus, the work done is positive