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kkurt [141]
4 years ago
13

Use the terms "force", "weight", "mass", and "inertia" to explain why it is easier to tackle a 220 lb football player than a 288

lb football player.
Physics
2 answers:
Tomtit [17]4 years ago
7 0
<span><u>Answer </u>
The mass of 220 lb football has less than 288 lb football. So, it will be easier to move it since it will require less force. The heavy football will have a bigger momentum. Since 288 lb has more weight than 220 lb, it will have bigger inertia making it difficult for the players to stop it.
This makes it easier to tackle 220 lb football than 288 lb football. 
</span>
Flura [38]4 years ago
6 0

There are several scientific aspects that require to be comprehended beforehand;

Force = mass * acceleration

Weight = mass * gravity (10m/s)

Momentum = mass * velocity

Inertia is a property of mass which is the tendency of an object to remain at constant velocity unless acted upon by an outside force.

Based on these formulae, it is evident that if the 220lb player and 288lb player are moving at the same velocity, the 288lb player has more inertia and momentum than the 22olb. Therefore, it would require more force (hence energy) to change the direction or the speed (in tackling) of the player with 288lb compared with what would be required on the 220lb player.

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Answer:

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Explanation:

The partial negative charge on oxygen and partial positive charge on hydrogen enables them to make hydrogen bond and also makes it to dissolve the the other substances having partial charges.

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A national college researcher reported that 67%of students who graduated from high school in 2012 enrolled in college. Twenty-ni
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Explanation:

Theorem of Binomial Distribution will apply here.

n = 29 , p = .67 , q = 0.33

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How do we find an atom's mass number?
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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
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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,

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} } }

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