Why do the passengers in high-altitude jet planes feel the sensation of weight while passengers in an orbiting space vehicle, su
1 answer:
Passengers in an aircraft are subject to the Normal and Gravity Force acting on them at a low 'orbit', so tiny that it can be many times compared to the same surface of the earth when speaking in general terms.
In a high orbit space vehicle or in the same space, said force decreases considerably or simply disappears, generating the sensation of weightlessness.
Remember that the Force of Gravity is given under the principle
Where,
G = Gravitational Universal constant
M = Mass of the planet
m = mass of the object
r = Distance from center of the planet
When the radius grows considerably the gravitational force begins to decrease.
You might be interested in
The box has 3 forces acting on it:
• its own weight (magnitude <em>w</em>, pointing downward)
• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)
• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)
Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have
• net parallel force:
∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>
• net perpendicular force:
∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0
Solve the net perpendicular force equation for the normal force:
<em>n</em> = <em>w</em> cos(35°)
<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)
<em>n</em> ≈ 120 N
Solve for the mag. of friction:
<em>f</em> = <em>µ</em> <em>n</em>
<em>f</em> = 0.25 (120 N)
<em>f</em> ≈ 30 N
Solve the net parallel force equation for the acceleration:
-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>
<em>a</em> ≈ (54.3157 N) / (15 kg)
<em>a</em> ≈ 3.6 m/s²
Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:
<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>
<em>v</em>² = 2 (3.6 m/s²) (3 m)
<em>v</em> = √(21.7263 m²/s²)
<em>v</em> ≈ 4.7 m/s
Answer:
✓ A cyclone device accumulates fine particulates from the air by making a dirty air stream flow in a spiral path inside a cylindrical chamber.
✘ It consists of several long and narrow fabric filter bags suspended upside-down in a large enclosure.
✓ When dirty air enters the chamber, the larger particulates strike the chamber wall and fall into a conical dust hopper at the bottom.
✘ Fans blow dirt-filled air upward from the bottom of the enclosure, trapping dirt particles inside the filter bags and releasing clean air from the top.
✓ The top of the chamber has an outlet that lets out cleaned air.
Basically, any of these choices that have the word "filter" are wrong. The point of the cyclone device is to separate the particles without the use of filters. You can tell the right answers based on the picture attached below.
