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Lyrx [107]
3 years ago
15

What type of research study is best for placing objects into categories? A. comparison B. relational C. simulation D. cause and

effect
Physics
1 answer:
myrzilka [38]3 years ago
7 0

Relational is the best type of research study for placing objects into categories.

Answer: Option B

<u>Explanation: </u>

Categorization is the process of making a set of objects having similar properties under a single category. The research study can be classified as quantitative, qualitative and descriptive types.

The categorisation is based on the qualitative research under the sub type of relation or classification research. Based on the relation between each objects, they will be grouped together.

If two or more objects have any kind of positive relations like similar behaviours, they will be grouped together and if their properties are different then the objects will be placed in different groups.

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What is the correct formula for power? A. Power = work / time B. Power = work * time C. Power = force * distance D. Power = work
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Answer:

A. Power = Work / Time

Explanation:

Power is the amount of work done over time, or rather the rate of work, which is given by the unit of watts (W). Since work is defined by Force * Displacement, we can also say Power = Force * Displacement / Time.

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A projectile has an initial horizontal velocity of 34.0 M/s at the edge of a roof top. Find the horizontal and vertical componen
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Answer:

v_x=34 m/s

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<u>Horizontal Launch</u>

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vx=v

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3 years ago
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
Crank

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.

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t=\frac{v_0}{g}

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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.

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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,

t_{tot}=2\dfrac{v_0}{g}+t_0

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H=v_0t_0+\dfrac{1}{2}gt_0^2

we solve for t_0 using the quadratic formula and take the positive value to get:

t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

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t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH}  }{g}}

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

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