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3 years ago

6

In a model of the solar system that was about as large as a college campus how large would the Sun be?

1 answer:

nordsb [41]3 years ago

8 0

Small car is the answer

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To transfer the visitor's city map into a two dimensional map,

Dovator [93]

Answer:

Imagine you have an orange. This is your imaginary Earth. When you look at it in any direction, you won’t be able to see all sides of it. But when you peel the orange, flatten and stretch it out, you can begin to see everything.

Similarly, a map projection is a method by which cartographers translate a sphere or globe into a two-dimensional representation. In other words, a map projection systematically renders a 3D ellipsoid (or spheroid) of Earth to a 2D map surface.

Explanation:

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1 year ago

At which of the following angles will the sunlight received at a location on Earth spread out over the largest area?

yarga [219]

10 degrees, because 0 degrees will receive the most sunlight, and 10 degrees is closest to 0 degrees.

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3 years ago

Read 2 more answers

What are the two <br>factors in which weight of object depends?

insens350 [35]

Answer:

An object's weight depends on its mass (the amount of matter it consists of) and the strength of the gravitational pull.

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3 years ago

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A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in 1.90 s. You may ignore air

Gekata [30.6K]

Answer: Height of building = 17.69m, velocity of brick = 18.6m/s

Explanation: From the question, the body has a zero initial speed, thus initial velocity (u) all through the motion is zero.

By ignoring air resistance makes it a free fall motion thus making it to accelerate constantly with a value of $a = 9.8m/s^{2}$ .

Time taken to fall = 1.90s

a)

thus the height of the building is calculated using the formulae below

$H = ut + \frac{1}{2} gt^{2}$

but u = 0 , hence

$H = \frac{1}{2} gt^{2}$

$H = \frac{1}{2} *9.8* 1.9^{2} \\\\H = 17.69m$

b)

to get the value of velocity (v) as the brick hits the ground, we use the formulae below

$v^{2} = u^{2} + 2aH$

but u= 0, hence

$v^{2} = 2gH\\$

$v^{2} = 2 * 9.8 * 17.69\\v = \sqrt{2 *9.8* 17.69} \\v = 18.62m/s$

find the attachment in this answer for the accleration- time graph, velocity- time graph and distance time graph

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4 years ago

Calculate the GPE of a rock with a mass of 55kg on a cliff that is 27m high

GaryK [48]

Answer:

<em>P=mgh</em>

<em>P=mghm=55</em>

<em>P=mghm=55g=9.8 or ~10</em>

<em>P=mghm=55g=9.8 or ~10h=27</em>

Explanation:

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4 0

2 years ago