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Anika [276]
3 years ago
6

these are the types of attractions found between molecules. their strength determines the state of the substance at room tempera

ture.
Physics
2 answers:
Elina [12.6K]3 years ago
7 0
<h2>Answer</h2>

The physical state of the elements depends upon the <u>attraction forces </u>and their <u>kinetic energy</u>.

<h2>Explanation</h2>

The elements or substances are fixed with each other with the help of different chemical forces including ionic bonding, covalent bonding, H- bonding etc. The strength of these forces is also one of the factors that affect their physical natures. For example, covalent or ionic bonds are the strongest bonds than all other bonds and metals that contain these forces are mostly in solid form. The kinetic motion of electrons in the element also affects the physical state of the element and potential of bonding.

insens350 [35]3 years ago
5 0

Answer:

Intermolecular forces

Explanation:

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PLEASE HELP ME IF YOU KNOW STUFF ABOUT ENERGY TYSM
Tomtit [17]

Answer:

  1. Kinetic Energy
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3 years ago
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Identifying the guilty party was mainly based on eyewitness accounts during what time period?
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The time period for guilty party was between 1900-1988.

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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
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.

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,

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

where t_0 is the time it takes the rock to go from the roof of the building to the ground, and it is given by

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}

Therefore the total time is

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}}

Now the average velocity is

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

v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }

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

5 0
3 years ago
traveling at about 30 mph, how many feet will the average driver cover from the time they see the danger until they hit the brak
barxatty [35]

Answer: 75 ft

Explanation:

Breaking distance = Speed²/ 20

= 30²/20

= 45 feet

Stopping distance = Speed + braking distance

= 30 + 45

= 75 ft

5 0
2 years ago
If a car accelerates at a uniform 4.0 m/s, how long will it take to reach a speed of 36.0 m/s,
11111nata11111 [884]

Answer:

9s

Explanation:

v=u+at

36=0+4t

t=36-0/4

t=9s

5 0
3 years ago
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