All categories

3 years ago

If a scale of the solar system was built where 1 mm equaled 1 mile, could the model be practically built in a city?

Physics

2 answers:

Y_Kistochka [10]3 years ago

7 0

No, it couldn't be.
On that scale, Neptune would be almost 1,740 MILES from the sun.
ON THAT SCALE !

Irina18 [472]3 years ago

3 0

Answer:

Explanation:

Given the relative scale (1mm = 1mile) lets calculate the size of the Sun and distance of planets from the Sun in mm.

Diameter of Sun = 864938 miles = 864938 mm = 864.938 meters

Distance of Earth from the Sun = 92955810 miles = 92955810 mm = 92.955 km = 57.75 miles

Distance of Jupiter from Sun = 480000000 miles = 480000000 mm = 480 km = 298 miles

As can be seen, the magnitude of scaled down distances is too huge to fit inside a city.

It will not be practically possible to make such a model.

You might be interested in

If wheel turning at a constant rate completes 100 revolutions in 10 s its angular speed is:

GenaCL600 [577]

The frequency of the wheel is given by:
$f= \frac{N}{t}$
where N is the number of revolutions and t is the time taken. By using N=100 and t=10 s, we find the frequency of the wheel:
$f= \frac{100}{10 s}=10 s^{-1}$

And now we can find the angular speed of the wheel, which is related to the frequency by:
$\omega=2 \pi f=2 \pi (10 s^{-1})=62.8 s^{-1}$

6 0

3 years ago

The speed of light in a solid is 1.24 x 108 m/s. <br> Calculate the index of refraction

Dahasolnce [82]

Answer:

125

Explanation:

7 0

2 years ago

In what direction is the weight vector always drawn?

Varvara68 [4.7K]

Yo sup??

The weight vector is usually drawn vertically downwards from the centre of the body.

It can be respectively resolved as well.

Hope this helps

3 0

2 years ago

Michael is playing with two horseshoe magnets. He is trying to get them to touch, but they will not regardless of how hard he tr

Nesterboy [21]

Since like poles repel, the two horseshoe magnets have like poles facing each other, hence they repel each other and therefore they will not come in contact

8 0

2 years ago

Read 2 more answers

A ball is kicked horizontally at 4.6 m/s off of a cliff 13.4 m high. How far from the cliff will it land.

Mekhanik [1.2K]

<h3>Answer:</h3>

7.53 m

<h3>Explanation:</h3>

<u>We are given:</u>

Initial Horizontal Velocity of the Ball = 4.6 m/s

Initial Vertical Velocity of the Ball = 0 m/s

Height from which ball is kicked = 13.4 m

<u>Time taken by the ball to reach the ground:</u>

The ball has an initial vertical velocity of 0 m/s

it also has a downward acceleration of 10 m/s² due to gravity

<u>Solving for the time taken:</u>

s = ut + 1/2(at²) [second equation of motion]

replacing the values

13.4 = (0)(t) + 1/2 (10)(t²)

13.4 = 5t²

t² = 13.4/5 [dividing both sides by 5]

t² = 2.68

t = 1.637 seconds [taking the square root of both sides]

<u>Horizontal distance covered by the ball:</u>

Since there are no horizontal opposing forces on the ball,

the ball will more horizontally at a velocity of 4.6 m/s until it hits the ground

We calculated that the ball will hit the ground in 1.637 seconds

<u>Distance covered:</u>

s = ut + 1/2 (at²) [seconds equation of motion]

s = ut [since a = 0m/s² in the horizontal plane]

replacing the values

s = 4.6 * 1.637

s = 7.53 m

Hence, the ball landed 7.53 m from the cliff

5 0

2 years ago