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AURORKA [14]
3 years ago
10

a scale model of the solar system where 50 cm represents 1.0x10 to the fifth km is actual distance what would be the dimension o

f the distance of Mars from the Sun?
Physics
1 answer:
Fofino [41]3 years ago
4 0

The distance between Mars and the Sun in the scale model would be 1140 m

Explanation:

In this scale model, we have:

x_1 = 50 cm represents an actual distance of

d_1 = 1.0\cdot 10^5 km

The actual distance between Mars and the Sun is 228 million km, therefore

d_2=228\cdot 10^6 km

On the scale model, this would corresponds to a distance of x_2.

Therefore, we can write the following proportion:

\frac{x_1}{d_1}=\frac{x_2}{d_2}

And solving for x_2, we find:

x_2=\frac{x_1 d_2}{d_1}=\frac{(50)(228\cdot 10^6)}{1\cdot 10^5}=1.14\cdot 10^5 cm = 1140 m

Learn more about distance:

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4) D: 84.84 metres

Explanation:

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Plugging in the relevant values, we have;

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Explanation:

According to the described situation we have the following data:

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Ferdinand's initial velocity: V_{o}=5 m/s

Time it takes a jump: t=0.6 s

We need to find the angle \theta at which Ferdinand jumps.

In order to do this, we first have to find the <u>horizontal component (or x-component)</u> of this initial velocity. Since we are dealing with parabolic movement, where velocity has x-component and y-component, and in this case we will choose the x-component to find the angle:

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