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just olya [345]
3 years ago
12

Why do the passengers in high-altitude jet planes feel the sensation of weight while passengers in an orbiting space vehicle, su

ch as a space shuttle, do not?
Physics
1 answer:
kykrilka [37]3 years ago
7 0

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

F_g = \frac{GMm}{r^2}

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.

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As an object moves, the distance it travels increases with time.<br><br> Agree<br> Disagree
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4 0
3 years ago
Read 2 more answers
From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. As the
Ivahew [28]

Answer:

The launching point is at a distance D = 962.2m and H = 39.2m

Explanation:

It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.

X axis

           x = Vox t

           t = x / vox

           t = 7.1 / 340

           t = 2.09 10-2 s

In this same time the height of the window fell

           Y = Voy t - ½ g t²

Let's calculate the initial vertical speed, this speed is in the window

           Voy = (Y + ½ g t²) / t

           Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209

            Voy = 27.7 m / s

We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s

             Vy = Voy - gt₂

             Vy = 0 -g t₂

             t₂ = Vy / g

             t₂ = 27.7 / 9.8

             t₂ = 2.83 s

This is the time it also takes to travel the horizontal and vertical distance

            X = Vox t₂

            D = 340 2.83

            D = 962.2 m

           

            Y = Voy₂– ½ g t₂²

            Y = 0 - ½ g t2

            H = Y = - ½ 9.8 2.83 2

            H = 39.2 m

The launching point is at a distance D = 962.2m and H = 39.2m

6 0
3 years ago
Plss help
Genrish500 [490]

Answer:

b

Explanation:

imagine urself on an elevator dont you feel lighter

3 0
3 years ago
Observe: Air pressure is equal to the weight of a column of air on a particular location. Airpressure is measured in millibars (
lesya [120]

Answer:

a) When moving towards a high pressure center the pressure values ​​increase in the equipment

b) This area is called high prison since the weight of the atmosphere on top is maximum

Explanation:

A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values ​​85.5 kPa.

When moving towards a high pressure center the pressure values ​​increase in the equipment

B) This area is called high prison since the weight of the atmosphere on top is maximum

in general they are areas of good weather

6 0
3 years ago
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