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3 years ago

I’m begging you please please ASAP ASAP please ASAP thank you please please ASAP

Mathematics

1 answer:

kozerog [31]3 years ago

6 0

Answer:

The ratio is $\frac{2}{3}$ .

Step-by-step explanation:

The ratio of the perimeters of Quad ABCD to Quad WXYZ = $\frac{perimeter of ABCD}{perimeter of WXYZ}$

But considering sides AB and WX,

representative factor for both figures = $\frac{12}{8}$

So that;

WX = 12

XY = 1.5 x 6 = 9

YZ = 1.5 x 7 = 10.5

WZ = 1.5 x 7 = 10.5

Thus,

perimeter of Quad ABCD = 6 + 7 + 7 + 8

= 28

perimeter of Quad WXYZ = 9 + 10.5 + 10.5 + 12

= 42

The ratio of the perimeters of Quad ABCD to Quad WXYZ = $\frac{28}{42}$

= $\frac{2}{3}$

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Answer:

$\boxed{\sf distance \: between \: the \: objects \: \: is \: 2.5 \: cm}$

Step-by-step explanation:

According to universal law of gravitation every object in the universe attracts every other particles that surrounds it with a force which is inversely proportional to the square of their distance of separation & directly proportional to

the product of their masses, given by standard formula.

$A = G \cdot \frac{m_1.m_2}{ {d}^{2} }$

where A,d & G are force of attraction, distance of separation & proportionality constant respectively.

Given:

A1 = 2 units

d1= 5 cm

A2 = 8 units

To find:

Distance of separation when the force of attraction is 8 units d2 = ?

Solution:

Substituting the given values in above at each point,

$A_1 = G \cdot \frac{m_1.m_2}{ {d_1}^{2} } \\ 2 = G \cdot \frac{m_1.m_2}{ {5}^{2} } \\ \sf similarly \: at \: second \: point \: of \: attraction \\ A_2 = G \cdot \frac{m_1.m_2}{ {d_2}^{2} } \\ 8 = G \cdot \frac{m_1.m_2}{ {d_2}^{2} } \\ \sf \: \: dividing \: both \: of \: the \: equation \\ \frac{2}{8} = \frac{G \cdot \frac{m_1.m_2}{ {5}^{2} }}{G \cdot \frac{m_1.m_2}{ {d_2}^{2} } } \\ \sf \: G \: m_1 and \: m_2 \: are \: \: constant \: hence \\ \sf they \: can \: be \: cancelled \: out \\ \frac{2}{8} = \frac{ \frac{1}{ {5}^{2} } }{ \frac{1}{ {d_2}^{2} } } \\ \sf rearranging \: above \: equation \\ \frac{1}{4} = \frac{{d_2}^{2} }{ {5}^{2} } \\ {d_2}^{2} = \frac{ {5}^{2} }{4 } \\ {d_2}^{2} = \frac{25}{4} \\ {d_2}^{2} = 6.25 \\ \sqrt{{d_2}^{2} } = \sqrt{6.25} \\ \boxed{ \sf{d_2} = 2.5 \: cm}$

Answer: the distance between the two objects is 2.5 cm, if the attraction between them is 8 units.

Learn more about universal law of gravitation here brainly.com/question/27244479

Thanks for joining brainly community!

7 0

2 years ago