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MakcuM [25]
3 years ago
5

Line segment Q R , Line segment R S and Line segment S Q are midsegments of Î"WXY. Triangle R Q S is inside triangle X Y W. Poin

t R is the midpoint of side X Y, point S is the midpoint of side Y W, and point Q is the midpoint of side X W. The length of Q R is 2. 93 centimeters, the length of R S is 2. 04 centimeters, and the length of Q S is 2. 28 centimeters. What is the perimeter of Î"WXY? 11. 57 cm 12. 22 cm 12. 46 cm 14. 50 cm.
Physics
1 answer:
Whitepunk [10]3 years ago
8 0

The perimeter of ΔWXY is : ( D ) 14.5 cm

<u>Calculating the </u><u>perimeter </u><u>of ΔWXY</u>

QR = WY / 2

RS = XW / 2

QS = XY / 2

Given that : QR = 2.93 cm ,  RS = 2.04 cm,  QS = 2.28 cm

Therefore

Perimeter of ΔWXY = ∑ WY + XW + XY  

                                 = 2SR + 2QS + 2QR

                                 = 2(2.04) + 2(2.28) + 2(2.93)

                                 = 14.5 cm

Hence we can conclude that the perimeter of ΔWXY = 14.5 cm

learn more about perimeter calculations : brainly.com/question/24744445

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Answer:

The object will have an upward acceleration

Explanation:

Let's consider the forces applied on the box. We have only two forces:

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Therefore, the net force on the box is:

F_{net}=F_p -W

Here, we know that force applied is equal or greater than the weight, so

F_p \geq W

And therefore the net force is greater than zero:

F_{net}\geq 0 (1)

According to Newton's second law of motion, the net force on the box is equal to the product between its mass and its acceleration:

F_{net}=ma

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m is the mass of the box

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Therefore, the box will have an upward acceleration.

In this case force example we have:

F_p = 100 N\\W = 40 N

So the mass of the box is

m=\frac{W}{g}=\frac{40}{10}=4 kg

So the net force is

F_{net}=F_p-W=100-40=60 N

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The charge of the sphere q = 46\times1.6\times10^{-19} C

The magnitude of the repulsive force between the charges pushing them a part is

Using coulomb law

F = \dfrac {kq_{1}q_{2}}{r^{2}}

F = \dfrac{9\times10^{9}\times (46\times1.6\times10^{-19})^{2}C^{2}}{2\times(5.90\times10^{-15})^{2}\ m^{2}}

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