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muminat
3 years ago
7

Find the dimension of the subspace of all vectors in set of real numbers R Superscript 4 whose first and third entries are equal

.
Mathematics
1 answer:
USPshnik [31]3 years ago
6 0

Answer with explanation:

Consider a vector having 4 elements

     \left[\begin{array}{cccc}a_{1}&a_{2}&a_{3}&a_{4}\end{array}\right]

it's first and third elements are equal.

That is,

     a_{1}=a_{4}

\left[\begin{array}{cccc}a_{1}&a_{2}&a_{3}&a_{4}\end{array}\right]\\\\R_{1}\rightarrow R_{1}-R_{3} \\\\\text{or}\\\\R_{3}\rightarrow R_{3}-R_{1}\\\\\left[\begin{array}{cccc}0&a_{2}&a_{3}&a_{4}\end{array}\right]\text{or}\left[\begin{array}{cccc}a_{1}&a_{2}&0&a_{4}\end{array}\right]

So, Dimension of vector=3

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A projectile is fired with muzzle speed 220 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe
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Step-by-step explanation:

For the purpose of this problem, we assume ballistic motion over a stationary flat Earth under the influence of gravity, with no air resistance.

We can divide the motion into two components, one vertical and one horizontal. For muzzle speed s and launch angle θ, the horizontal speed is presumed constant at s·cos(θ). The initial vertical speed is then s·sin(θ) and the (x, y) coordinates as a function of time are ...

  (x, y) = (s·cos(θ)·t, -4.9t² +s·sin(θ)·t + h₀) . . . . . where h₀ is the initial height

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Using the quadratic formula, we find t at the time of landing to be ...

  t = (-s·sin(θ) - √((s·sin(θ))²-4(-4.9)(h₀)))/(2(-4.9))

  t = (s/9.8)(sin(θ) +√(sin(θ)² +19.6h₀/s²))

For s = 220, θ = 45°, and h₀ = 30, the time of flight is ...

  t ≈ 31.939 seconds

Then the horizontal travel is

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__

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  landing speed = s·√(cos(θ)² +sin(θ)² +19.6h₀/s²) ≈ s√(1 +0.012149)

  ≈ 221.33 m/s

_____

Note that the landing speed represents the speed the projectile has as a consequence of the potential energy of its initial height being converted to kinetic energy that adds to the kinetic energy due to its initial muzzle velocity.

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