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PtichkaEL [24]
2 years ago
11

of 5.25 newtons acts on an object of unknown mass at a distance of 6.9 x 10^8 meters from the center of Earth. To increase the f

orce to 2.5 times its original value, how far should the object be from the center of Earth?
Mathematics
1 answer:
schepotkina [342]2 years ago
8 0
Hey

The nice thing about this question is that you don't actually need to worry about numbers until the very end! It's just a matter of setting up the right equation. First of all, remember Newton's law of universal gravitation:<span>F=<span><span>G<span>m1</span><span>m2</span></span><span>r2</span></span></span>Where F is the force of gravity between the two objects, G is the gravitational constant, m(1) is the mass of one object, m(2) is the mass of the other object, and r is the distance between the two objects. We can rearrange this equation to solve for G instead (you'll see why this is important soon):<span>G=<span><span>F<span>r2</span></span><span><span>m1</span><span>m2</span></span></span></span>We have two situations here: (a) (the starting position of the objects) and (b) (the final position of the objects). We can use Newton's Law of gravitation to describe the force of gravity between these two objects during situations (a) and (b), but we can actually set these equations equal to one another because the value of G will be the same in both (it's a constant!) - now you see why I rearranged the equation above.<span><span><span><span>Fa</span><span>r2a</span></span><span><span>m<span>1a</span></span><span>m<span>2a</span></span></span></span>=<span><span><span>Fb</span><span>r2b</span></span><span><span>m<span>1b</span></span><span>m<span>2b</span></span></span></span></span>I know this equation looks a bit messy, so let's simplify is a bit. The masses of the objects are the same in both situations (we're dealing with the unknown object and the Earth both times), so we can take both those guys out of the equation:<span><span><span>Fa</span><span>r2a</span></span>=<span><span>Fb</span><span>r2b</span></span></span>Already it's looking better! We also know we want the force in the situation (b) to be 2.5 times the force in the situation (a). In other words, F(b) = 2.5F(a). We can plug that into the equation, and then solve becomes very easy:<span><span><span>Fa</span><span>r2a</span></span>=<span>2.5<span>Fa</span><span>r2b</span></span></span><span><span>r2a</span>=<span>2.5<span>r2b</span></span></span><span><span>rb</span>=<span><span>ra</span><span>2.5<span>−−−</span>√</span></span></span><span><span>rb</span>=<span><span>6.9×<span>108</span></span><span>2.5<span>−−−</span>√</span></span></span><span><span>rb</span>=4.4×<span>108</span></span><span> So the new distance would have to be 4.4 x 10^8 m for the force to be 2.5 times as great. Notice how the fact that the original force was 5.25 N doesn't actually matter! And this answer also makes sense because you would expect the gravitational force to become stronger as you move the objects closer together. If you have any questions please let me know!</span>

Hoped I Helped
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How to do this problem??
lorasvet [3.4K]
We know that the sum of the inner angles of any triangle is 180º

72º + (7x + 3)º + (3x + 5)º = 180º
72º + 7xº + 3º + 3xº + 5º = 180
7xº + 3xº = 180º - 72º - 3º - 5º
10xº = 100º
x^0 =  \frac{100^0}{10^0}
x^0 = 10\to\:\boxed{\boxed{x = 10^0}}\end{array}}\qquad\quad\checkmark

The sum of the external angle (9y + 1)º with inner angle (3x + 5) = 180 °, <span>Replace the measure of "x" found:
</span>
(9y + 1)º + (3x + 5)º = 180º
9yº + 1º + 3xº + 5º = 180º
9yº + 1º + 3.(10)º + 5º = 180º
9yº + 1º + 30º + 5º = 180º
9yº = 180º - 1º - 30º - 5º
9yº = 144º
y^0 =  \frac{144^0}{9^0}
y^0 = 16\to\:\boxed{\boxed{y = 16^0}}\end{array}}\qquad\quad\checkmark

Answer:
<span>The measures ​​of "x" and "y" are respectively: 10º and 16º</span>
6 0
2 years ago
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2 years ago
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gregori [183]

Answer:

p = 5 and b = 8

Step-by-step explanation:

First, create a system of equations:

2p + 4b = 42

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Solve by elimination by multiplying the top equation by -7 and the bottom equation by 2:

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4 0
2 years ago
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(easy) If ΔEFG ~ ΔLMN with a ratio of 3:1, which of the following is true?
alexandr402 [8]

Answer:

segment EG over segment LN equals segment FG over segment MN

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its corresponding sides is proportional and its corresponding angles are congruent

In this problem

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EF and LM

EG and LN

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The corresponding angles are

∠E≅∠L

∠F≅∠M

∠G≅∠N

therefore

EF/LM=EG/LN=FG/MN=3/1

7 0
3 years ago
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Novay_Z [31]

814,180. Hope I helped. If i didn't I am sos sorry. I would take 7.9 or .079 to 754,844 to get your number.

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