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Juliette [100K]
3 years ago
7

Space Station Suppose a space-station is designed in s shape of a torus such as the one depicted in Stanley Kubrick's "2001: A s

pace Odyssey". Suppose the space station has a diameter of 175 meters. At what rotational rate should the sta- tion be spun so that occupants standing on the inside surface of the outer wall (maximum distance from the center) experience earth-normal artificial gravity? Give your answer in terms of rotations per minute Suppose that inside the space station there is a transport cart for transport around the rim that can reach a maxumum speed of 25.0 meters per second. Using such a cart, how could you tell which way the space station from spinning from inside without looking out any windows? Be quite specific about what you experience when you drive around with the cart in one direction or the other
Physics
1 answer:
yaroslaw [1]3 years ago
3 0

Answer:

w = 3.2 rev / min

Explanation:

For this exercise we will use the centrine acceleration equal to the acceleration of gravity

      a = v² / r

Angular and linear variables are related.

     v = w r

Let's replace

     a = w² r = g

     w = √ g / r

     r = d / 2

     r = 175/2 = 87.5 m

    w = √( 9.8 / 87.5)

    w = 0.3347 rad / s

Let's reduce to rotations per min

     w = 0.3347 rad / s (1 rov / 2pi rad) (60 s / 1 min)

     w = 3.2 rev / min

Suppose the space station rotates counterclockwise, we have two possibilities for the car

The first car turns counterclockwise (same direction of the station

     v_{c} =  w_{c} r

     [texwv_{c}[/tex] =  v_{c} / r

     [texwv_{c}[/tex] = 25.0 / 87.5

     [texwv_{c}[/tex] = 0.286 rad / s

When the two rotate in the same direction their angular speeds are subtracted

     w total = w -[texwv_{c}[/tex]

     w total = 0.3347 - 0.286

    w  total= 0.487 rad / s

The car goes in the opposite direction of the station the speeds add up

    w = 0.3347 + 0.286

    w = 0.62 rad / s

From this values ​​we can see that the person feels a variation of the acceleration of gravity, feels that he has less weight when he goes in the same direction of the season and that his weight increases when he goes in the opposite direction to the season.

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Two geological field teams are working in a remote area. A global positioning system (GPS) tracker at their base camp shows the
Masja [62]

Answer:

distance of 2nd team from 1st team will be:  58.2

Direction of 2nd team from 1st team will be:  14.90 deg North of east

Explanation:

ASSUME Vector is R and  makes angle A with +x-axis,

therefore component of vector R is

R_x = Rcos A

R_y = Rsin A

From above relation

Assuming base camp as the origin, location of 1st team is

R_1 = 37 km away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)

R_{1x} = -R_1*cos A_1 = -37*cos 21 deg = -34.54 km

R_{1y} = R_1*sin A_1 = 37*sin 21 deg = 13.25 km

location of 2nd team is at

R_2 = 32 km, at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)

R_{2x} = R_2*cos A_2 = 32*cos 58 deg = 16.95 km

R_{2y} = R_2*sin A_2 = 32*sin 58 deg = 27.13 km

Now position of 2nd team with respect to 1st team will be given by:

R_3 = R_2 - R_1

R_3 = (R_{2x} - R_{1x}) i + (R_{2y} - R_{1y}) j

Using above values:

R_3 = (16.95 - (-34.54)) i + (27.13 - 13.42) j

R_3 = 51.49 i + 13.71 j

distance of 2nd team from 1st team will be:

\left | R_3 \right | = \sqrt (51.49^2 +13.71^2)

\left | R_3 \right | = 53.28 km = 58.2 km

Direction of 2nd team from 1st team will be:

Direction = tan^{-1} \frac{R_{3y}}{R_{3x}} = tan^{-1}[ \frac{13.71}{51.49}]

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A slab of glass 8.0 cm thick is placed upon a printed page. If the refractive index of the glass is 1.5, how far from the surfac
7nadin3 [17]

Answer:

5.3 cm

Explanation:

This question is an illustration of real and apparent distance.

From the question, we have the following given parameters

Real Distance, R = 8.0cm

Refractive Index, μ = 1.5

Required

Determine the apparent distance (A)

The relationship between R, A and μ is:

μ = R/A

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Refractive Index = Real Distance ÷ Apparent Distance

Substitute values in the above formula

1.5 = 8/A

Multiply both sides by A

1.5 * A = A * 8/A

1.5A = 8

Divide both side by 1.5

1.5A/1.5 = 8/1.5

A = 8/1.5

A = 5.3cm

Hence, the letters would appear at a distance of 5.3cm

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