2 years ago

Find the inertia tensor for an equilateral triangle in the xy plane.

1 answer:

7 0

Find the inertia tensor for an equilateral triangle in the xy plane. Take the mass of the triangle to be M and the length of a side of the triangle to be b. Express your answer below as pure numbers in units of Mb^2. Place the origin on the midpoint of one side and set the y-axis to be along the symmetry axis.

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Find each product by distributing.<br> (2a - 5)(a? – 2a + 1)

Westkost [7]

Answer:

A = 1 , 1 , $\frac{5}{2}$

Step-by-step explanation:

Given equation as

(2a -5) (a² -2a +1) = 0

Or, (2a - 5) (a - 1 )² = 0

Or, (2a - 5) (a - 1) (a - 1) = 0

So, value of a are :

2a - 5 =0, a - 1 =0, a- 1 = 0

i.e a = $\frac{5}{2}$

and a = 1 , 1

Answer

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kolbaska11 [484]

Answer:

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Step-by-step explanation:

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