The forces acting on the elevator are:
Gravity force
Tension force
Air resistance
Explanation:
Let's go through each of the forces listed and see which ones are acting on the elevator.
- Normal force: NO. The normal force is a force exerted by a surface whenever there is another object "pushing" on it. For instance, when a box is at rest on a table, the box is "pushing" on the table (due to its weight), and the table "pushes back" on the box, upward, in order to balance its weight: this is the normal force. In this case, the elevator is lifted, so it is not pushing on anything, therefore there is no normal force.
- Gravity force: YES. The force of gravity acts on every object located in the gravitational field of the Earth; it pulls downward, and its magnitude is
, where m is the mass of the object and g is the acceleration of gravity. - Applied force: NO. Here there is no applied force, since there is nobody "pushing" or "pulling" the elevator.
- Friction force: NO. As we are considering the forces on the elevator, and the elevator is not sliding against any surfaces, there is no force of friction. (The force of friction acts whenever there are two surfaces sliding against each other, which is not the case here)
- Tension force: YES. The tension force is the force exerted by a rope or a string when pulling an object. In this case, there are four ropes pulling the elevator, therefore there are 4 forces of tension acting on the elevator, upward.
- Air resistance: YES. As the elevator is moving through the air, the interaction between the molecules of air with the surface of the elevator produces a force (called air resistance) that "resists" the motion of the elevator, therefore pushing downward. However, the magnitude of this force is negligible in this case.
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Answer:
a) 
b) 
Explanation:
given,
n =1.5 for glass surface
n = 1 for air
incidence angle = 45°
using Fresnel equation of reflectivity of S and P polarized light

using snell's law to calculate θ t


a) 

b) 

Answer:
t = 22.2 s
Explanation:
angular distance covered in the 36.0 s is
θ = ω(avg)t = ½(10.0 + 30.0)36 = 720 radians
720/2 = 360 radians
α = Δω/t = (30 - 10)/36 = 5/9 rad/s²
θ = ω₀t + ½αt²
360 = 10.0t + ½(5/9)t²
0 = (5/18)t² + 10.0t - 360
0 = t² + 36t - 1296
t = (-36 ±√(36² - 4(1)(-1296))) / 2
t = (-36 ±√(6480)) / 2
t = -18 ±√1620
we ignore the negative time result as it occurs before we care.
t = -18 + √1620 = 22.249223... s
The car heads east at an average speed of 50 miles per hour from the intersection point towards East. The truck heads east at an average speed of 60 miles per hour from the intersection point towards South.
The distance of car from the intersection point after t hours is
.
The distance of truck from the intersection point after t hours is
.
Since these distances are perpendicular to each other, distance apart d (in miles) at the end of t hours is

Thus the distance apart is 
Space debris that enters earths atmosphere