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3 years ago

Why is polaris used as a celestial reference point?

1 answer:

8 0

At the equator, the North Celestial and South Celestial Poles would lie on the horizon where the meridian intersects the horizon. Polaris is called the "North Star." ... Polaris is special because the Earth's North Pole points almost exactly towards it.

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A busy waitress slides a plate of apple pie along a counter to a hungry customer sitting near the

Ostrovityanka [42]

Answer:

a. 0.5307 sec

b. 0.4458 m

c. = $v_x=0.84\ m/s\ ,\ v_y=5.2\ m/s$

Explanation:

Horizontal Motion

It describes the dynamics of an object thrown horizontally in free air. The initial horizontal velocity is maintained all the time since no horizontal forces are acting. The initial vertical velocity is zero at launch time, but it grows downwards powered by the acceleration of gravity.

The object hits the ground at a distance x from the point of launching, after having traveled a vertical distance $y_o$ , taking a time t to complete the travel. The formulas who relate the different magnitudes are

$v_x=v_o$

The horizontal velocity $v_x$ is the same regardless of the elapsed time

$v_y=gt$

$x=v_ot$

$\displaystyle y=y_o-\frac{gt^2}{2}$

The plate of apple pie left the counter at a speed

$v_o=0.84\ m/s$

The counter is $y_o=1.38\ m$ high.

Knowing that

$y_o=1.38\ m$

We use this formula to compute t

$\displaystyle y=y_o-\frac{gt^2}{2}$

At the moment when the plate hits the floor y=0

$\displaystyle 0=y_o-\frac{gt^2}{2}$

$\displaystyle y_o=\frac{gt^2}{2}$

Solving for t

$\displaystyle t=\sqrt{\frac{2y_o}{g}}$

$\displaystyle t=\sqrt{\frac{2(1.38)}{9.8}}$

$t=0.5307\ sec$

$x=(0.84)(0.5307)$

$x=0.4458\ m$

$v_x=0.84\ m/s$

$v_y=(9.8)(0.5307)$

$v_y=5.2\ m/s$

3 0

4 years ago

Given that two vectors A = 5i-7j-3k, B = -4i+4j-8k find A×B

WARRIOR [948]

$\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}$

Explanation:

Given:

$\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}$

$\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}$

The cross product $\textbf{A}×\textbf{B}$ is given by

$\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|$

$= \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}$