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2 years ago

9

Goal posts at the ends of football fields are padded as a safety measure for players who might run into them. How does thick pad

ding around the goal post reduce injuries to players?

2 answers:

dimaraw [331]2 years ago

8 0

When looking at it from a physics standpoint, the thick padding decreases the amount of force that a player experiences. Force equals mass times acceleration where acceleration is distance over time. The thick padding increases the amount of time the player makes contact with the post which decreases force.

likoan [24]2 years ago

8 0

The padding around the goal post increases the time of the collision between the player and the post, which decreases the force exerted to bring the player to a stop.

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<h2>The hiker will go up to 850 m on the hill</h2>

Explanation:

The total energy gained by the hiker = 140 x 4186 J

This energy is consumed in the potential energy acquired , while climbing up the hill.

The potential energy P.E = mass of hiker x acceleration due to gravity x height

Thus

140 x 4186 = 69 x 10 x h

or h = $\frac{4186x140}{69x10}$ = 850 m

If the 20% of the total energy is used

the height h₀ = $\frac{0.2x4186x140}{69x10}$ = 170 m

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2 years ago

Read 2 more answers

During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s

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Answer:

part a : <em>The dry unit weight is 0.0616 </em> $lb/in^3$ <em />

part b : <em>The void ratio is 0.77</em>

part c : <em>Degree of Saturation is 0.43</em>

part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}$

Here

$\gamma_d$ is the dry unit weight which is to be calculated
γ is the bulk unit weight given as

$\gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3$

w is the moisture content in percentage, given as 12%

Substituting values

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3$

<em>The dry unit weight is 0.0616 </em> $lb/in^3$ <em />

Part b

Void Ratio

The void ratio is given as

$e=\frac{G_s \gamma_w}{\gamma_d} -1$

Here

e is the void ratio which is to be calculated

$\gamma_d$ is the dry unit weight which is calculated in part a
$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

Substituting values

$e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766$

<em>The void ratio is 0.77</em>

Part c

Degree of Saturation

Degree of Saturation is given as

$S=\frac{G w}{e}$

Here

e is the void ratio which is calculated in part b

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

$S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261$

<em>Degree of Saturation is 0.43</em>

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}$

Here

$\gamma_{zav}$ is the zero air unit weight which is to be calculated

$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\$

Now as the volume is known, the the overall weight is given as

$weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb$

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

4 0

2 years ago

Suppose that you purchased a water bed with the dimensions 2.55 m Ã 2.53 dm Ã 245 cm. what mass of water does this bed contain

Nadya [2.5K]

dimensions of the bed is given as

$length = 2.55 m$

$thickness = 2.53 dm = 0.253 m$

$width = 245 cm = 2.45 m$

now the volume of the bed is given as

$V = 2.55 * 0.253 * 2.45 m^3$

$V = 1.581m^3$

now the mass of water in it is given as

$mass = density * volume$

$mass = 1000* 1.581$

$mass = 1581 kg$

<em>so it will contain 1581 kg mass in it</em>

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2 years ago

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Answer:

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Explanation:

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Let's classify the structure from highest to lowest

F₂d> F₂a = F₂e> F₂c> F₂b = F₂f

I mean the combinations are

d> a = e> c> b = f

6 0

2 years ago