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2 years ago

A uniform brick of length 26 m is placed over the edge of a horizontal surface. (x=26 m)

Physics

1 answer:

DerKrebs [107]2 years ago

4 0

Answer:

Maximum 13 m overhang

Explanation:

Since the brick is uniform. Its mass must be equally distributed along its length. So the center of mass is in geometric center, which is 26 / 2 = 13 m from each end of the brick.

For the brick to not tip over, its center of mass must NOT be overhung over the edge. It stays just on the edge. So the maximum overhang is 13 m.

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Two vectors are given by a = 8.6i + 5.1 j and b = 931 + 9.5.

skelet666 [1.2K]

Answer:

Assuming b = 9.3i + 9.5j (b = 931 + 9.5 is wrong):

a) a×b = 34.27k

b) a·b = 128.43

c) (a + b)·b = 305.17

d) The component of a along the direction of b = 9.66

Explanation:

Assuming b = 9.3i + 9.5j (b = 931 + 9.5 is wrong) we can proceed as follows:

a) The vectorial product, a×b is:

$a \times b = (8.6*9.5 - 5.1*9.3)k = 34.27k$

b) The escalar product a·b is:

$a\cdot b = (8.6*9.3) + (5.1*9.5) = 128.43$

c) Asumming (a + b)·b instead a+b·b we have:

$(a + b)\cdot b = [(8.6 + 9.3)i + (5.1 + 9.5)j]\cdot (9.3i + 9.5j) = (17.9i + 14.6j)\cdot (9.3i + 9.5j) = 305.17$

d) The component of a along the direction of b is:

$a*cos(\theta) = \frac{a\cdot b}{|b|} = \frac{128.43}{\sqrt{9.3^{2} + 9.5^{2}}} = 9.66$

I hope it helps you!

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