Answer:
The value of tension on the cable T = 1065.6 N
Explanation:
Mass = 888 kg
Initial velocity ( u )= 0.8 
Final velocity ( V ) = 0
Distance traveled before come to rest = 0.2667 m
Now use third law of motion 
= 
- 2 a s
Put all the values in above formula we get,
⇒ 0 = 
- 2 × a ×0.2667
⇒ a = 1.2 
This is the deceleration of the box.
Tension in the cable is given by T = F = m × a
Put all the values in above formula we get,
T = 888 × 1.2
T = 1065.6 N
This is the value of tension on the cable.
Answer: apparent weighlessness.
Explanation:
1) Balance of forces on a person falling:
i) To answer this question we will deal with the assumption of non-drag force (abscence of air).
ii) When a person is dropped, and there is not air resistance, the only force acting on the person's body is the Earth's gravitational attraction (downward), which is the responsible for the gravitational acceleration (around 9.8 m/s²).
iii) Under that sceneraio, there is not normal force acting on the person (the normal force is the force that the floor or a chair exerts on a body to balance the gravitational force when the body is on it).
2) This is, the person does not feel a pressure upward, which is he/she does not feel the weight: freefalling is a situation of apparent weigthlessness.
3) True weightlessness is when the object is in a place where there exists not grativational acceleration: for example a point between two planes where the grativational forces are equal in magnitude but opposing in direction and so they cancel each other.
Therefore, you conclude that, assuming no air resistance, a person in this ride experiencing apparent weightlessness.
Given that,
Distance in south-west direction = 250 km
Projected angle to east = 60°
East component = ?
since,
cos ∅ = base/hypotenuse
base= hyp * cos ∅
East component = 250 * cos 60°
East component = 125 km
We need the power law for the change in potential energy (due to the Coulomb force) in bringing a charge q from infinity to distance r from charge Q. We are only interested in the ratio U₁/U₂, so I'm not going to bother with constants (like the permittivity of space).
<span>The potential energy of charge q is proportional to </span>
<span>∫[s=r to ∞] qQs⁻²ds = -qQs⁻¹|[s=r to ∞] = qQr⁻¹, </span>
<span>so if r₂ = 3r₁ and q₂ = q₁/4, then </span>
<span>U₁/U₂ = q₁Qr₂/(r₁q₂Q) = (q₁/q₂)(r₂/r₁) </span>
<span>= 4•3 = 12.</span>
When using the right-hand rule to determine the direction of the magnetic force on a charge, which part of the hand points in the direction that the charge is moving? The answer is <span>thumb.
</span>One way to remember this is that there is one velocity, represented accordingly by the thumb. There are many field lines, represented accordingly by the fingers. The force is in the direction you would push with your palm. The force on a negative charge is in exactly the opposite direction to that on a positive charge. Because the force is always perpendicular to the velocity vector, a pure magnetic field will not accelerate a charged particle in a single direction, however will produce circular or helical motion (a concept explored in more detail in future sections). It is important to note that magnetic field will not exert a force on a static electric charge. These two observations are in keeping with the rule that <span>magnetic fields do no </span>work<span>.</span>
