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Nimfa-mama [501]
2 years ago
13

Two similar pyramids have base areas of 12.2 cm2 and 16 cm2. the surface area of the larger pyramid is 56 cm2. what is the surfa

ce area of the smaller pyramid? 40.1 cm2 42.7 cm2 52.2 cm2 59.8 cm2
Physics
2 answers:
Jlenok [28]2 years ago
6 0

The surface area of the smaller pyramid is determined as 42.7 cm².

<h3>Area of pyramids</h3>

The base area of the two similar pyramid is given as follows;

Base area of large pyramid = 16 cm²

Base area of smaller pyramid = 12.2 cm²

<h3>Surface area of the similar pyramids</h3>

The surface area of the smaller pyramid is calculated as follows;

s₁/b₁ = s₂/b₂

s₁/12.2 = 56/16

s₁ = 42.7 cm²

Thus, the surface area of the smaller pyramid is determined as 42.7 cm².

Learn more about base area of pyramids here: brainly.com/question/16060915

#SPJ4

Verdich [7]2 years ago
3 0

Answer:

42.7

Explanation:

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An airplane is heading due south at a speed of 690 km/h . A) If a wind begins blowing from the southwest at a speed of 90 km/h (
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Answer:a) 629,5851 km/h in magnitude b)629,5851 km/h at 84,2 degrees from east pointing south direction or in vector form 626,6396 km/h south + 63,6396km/h  east. c) 16,5 km NE of the desired position

Explanation:

Since the plane is flying south at 690 km/h and the wind is blowing at assumed constant speed of 90 km/h from SW, we get a triangle relation where

 

see fig 1

Then we can decompose those 90 km/h into vectors, one north and one east, both of the same magnitude, since the angle is 45 degrees with respect to the east, that is direction norhteast or NE, then

90 km/h NE= 63,6396 km/h north + 63,6396 km/h east,

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90/\sqrt{2} km/h= 63,6396 km/h.  

Now the total speed of the plane is

690km/h south + 63,6396 km/h north +63,6396 km/h east,

this is 626,3604 km/h south + 63,6396 km/h east,  here north is as if we had -south.

then using again Pythagoras we get the magnitude of the total speed it is

\sqrt{626,3604 ^2+63,6396^2} km/h=629,5851km/h,

the direction is calculated with respect to the south using trigonometry, we know the

sin x= cathetus opposed / hypotenuse,

then

x= sin^{-1}'frac{63,6396}{629,5851}=5,801 degrees from South as reference (0 degrees) in East direction or as usual 84,2 degrees from east pointing south or in vector form

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Xdetour{east}= 63,6396 km/h* (11 min* h)/(60 min)=11,6672 km=Xdetour{north} ,

since 11 min=11/60 hours=0.1833 hours.

Then the total detour from the expected position, the one it should have without the influence of the wind, we get  

Xdetour=[/tex]\sqrt{2*  11,6672x^{2} }[/tex]  = 16,5km at 45 degrees from east pointing north

The situation is sketched as follows  see fig 2

 

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