The vector perpendicular to the plane of A = 3i+ 6j - 2k and B = 4i-j +3k is 16 i - 17 j - 27 k
Let r be the vector perpendicular to A and B,
r = A * B
A = 3i + 6j - 2k
B = 4i - j + 3k
a1 = 3
a2 = 6
a3 = - 2
b1 = 4
b2 = - 1
b3 = 3
a * b = ( a2 b3 - b2 a3 ) i + ( a3 b1 - b3 a1 ) j + ( a1 b2 - b1 a2 ) k
a * b = [ ( 6 * 3 ) - ( - 1 * - 2 ) ] i + [ ( - 2 * 4 ) - ( 3 * 3 ) ] j + [ ( 3 * - 1 ) - ( 4 * 6 ) ] k
a * b = 16 i - 17 j - 27 k
The perpendicular vector, r = 16 i - 17 j - 27 k
Therefore, the vector perpendicular to the plane of A = 3i + 6j - 2k and B = 4i - j + 3k is 16 i - 17 j - 27 k
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Answer:
A. The target nucleus split into two nuclei, each with fewer nucleons than the original.
Explanation:
Answer:
Its diameter increases as it flows down from the pipe. Assuming laminar flow for the water, then Bernoulli's equation can be applied.
P1-P2 + (rho)g(h1 - h2) + 1/2(rho)(v1² - v2²) = 0
Explanation:
P1 = P2 = atmospheric pressure so, P1 - P2 = 0
h1 is greater than h2 so h1-h2 is positive. Rearranging the equation above 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho = v2²
From the continuity equation for fluids
A1v1 = A2v2
v2 = A1v1/A2
Substituting into the equation above
(A1v1/A2)² = 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho
Making A2² the subject of the formula,
A2² = (A1v1)²× rho/(2{ (rho)g(h1-h2) + 1/2(rho)v1²}
The denominator will be greater than the numerator and as a result the diameter of the flowing stream decreases.
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