Answer:
the branch of mechanics concerned with the interaction of electric currents with magnetic fields or with other electric currents.
Explanation:
Answer:
<h2>1200 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 300 × 4
We have the final answer as
<h3>1200 N</h3>
Hope this helps you
Answer:
Air resistance/Drag (this is the proper term for air resistance) or gravity or friction
It depends on the direction and if you throw it across a surface or if you throw it straight upwards.
I hope this helps!!
The man is pulled upward by a net force of
∑ <em>F</em> = (86 kg) (1.35 m/s²) = 116.1 N
The net force is comprised of the tension <em>T</em> in the cable and the man's weight <em>W</em>, such that
∑ <em>F</em> = <em>T</em> - <em>W</em> = 116.1 N
<em>T</em> = (86 kg) (9.80 m/s²) + 116.1 N
<em>T</em> = 958.9 N ≈ 960 N
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
