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natali 33 [55]
3 years ago
12

Find the unit vector in the direction of vector B=4i+2j-5k​

Physics
1 answer:
solmaris [256]3 years ago
7 0

Answer:

\frac{4}{\sqrt{45} } i\,+\frac{2}{\sqrt{45} } j-\frac{5}{\sqrt{45} } k

Explanation:

for the unit vector, we need to divide the given vector by its norm, because it should be in the SAME direction as the original vector, but of magnitude "1".

We notice that the norm of the given vector is:

\sqrt{4^2+2^2+(-5)^2} =\sqrt{45}

Then, the unit vector becomes:

\frac{4}{\sqrt{45} } i\,+\frac{2}{\sqrt{45} } j-\frac{5}{\sqrt{45} } k

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A player kicks a football from ground level with a velocity of 26.2m/s at an angle of 34.2° above the horizontal. How far back f
Amanda [17]

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.

For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

<h3>Explanation</h3>

How long does it take for the ball to reach the goal?

Let the distance between the kicker and the goal be x meters.

Horizontal velocity of the ball will always be 26.2\times\cos{34.2\textdegree} until it lands if there's no air resistance.

The ball will arrive at the goal in \displaystyle \frac{x}{26.2\times\cos{34.2\textdegree}} seconds after it leaves the kicker.

What will be the height of the ball when it reaches the goal?

Consider the equation

\displaystyle h(t) = -\frac{1}{2}\cdot g\cdot t^{2} + v_{0,\;\text{vertical}} \cdot t + h_0.

For this soccer ball:

  • g = 9.81\;\text{m}\cdot\text{s}^{-2},
  • v_{0,\;\text{vertical}} = 26.2\times \sin{34.2\textdegree{}}\;\text{m}\cdot\text{s}^{-2},
  • h_0 = 0 since the player kicks the ball "from ground level."

\displaystyle t=\frac{x}{26.2\times\cos{34.2\textdegree}}

when the ball reaches the goal.

\displaystyle h= - 9.81 \times \frac{x^2}{(26.2\times\cos{34.2\textdegree})^2} + (26.2 \times \sin{34.2\textdegree})\times\frac{x}{26.2\times\cos{34.2\textdegree}} \\\phantom{h} = -\frac{9.81}{(26.2\times\cos{34.2\textdegree})^2}\cdot x^{2} + \frac{\sin{34.2\textdegree}}{\cos{34.2\textdegree}}\cdot x.

Solve this quadratic equation for x, x > 0.

  • x = 65.1 meters when h = 0 meters.
  • x = 6.54 or 58.5 meters when h = 4 meters.

In other words,

  • For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.
  • For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

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3 years ago
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Answer:

If there are no forces acting upon the object, then there is no acceleration. If there is no acceleration, then the object will move with a constant velocity.

Explanation:

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Gravitational Potential Energy = mgh

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