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2 years ago

⃗ has the magnitude 15 cm and it makes and angle 40° with positive X axis. Calculate the X and Y component of ⃗.

Physics

1 answer:

topjm [15]2 years ago

3 0

The X component of the vector is 11.5 while the Y component is 9.6.

<h3>What are the components of a vector?</h3>

We know that a vector must posses both magnitude and direction. Now the vector could be resolved into the vertical and the horizontal components of the mixture.

In resolving this vector;

X component = 15 cm cos 40 degrees = 11.5

Y component = 15 cm sin 40 degrees = 9.6

Thus the X component of the vector is 11.5 while the Y component is 9.6.

Learn more about vectors:brainly.com/question/13322477

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A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

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Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,