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Mamont248 [21]
3 years ago
12

A football player kicks a football downfield. The height of the football increases until it reaches a maximum height of 15 yards

, 30 yards away from the player. A second kick is modeled by f(x)=−0.032x(x−50), where f is the height (in yards) and x is the horizontal distance (in yards). Compare the distances that the footballs travel before hitting the ground.
Physics
2 answers:
Trava [24]3 years ago
8 0

Answer:

kick 1 has travelled 15 + 15 = 30 yards before hitting the ground

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards  

Explanation:

1st kick travelled 15 yards to reach maximum height of 8 yards

so, it has travelled 15 + 15 = 30 yards before hitting the ground

2nd kick is given by the equation

y (x) = -0.032x(x - 50)

Y = 1.6 X - 0.032x^2

we know that maximum height occurs is given as

x = -\frac{b}{2a}

y =- \frac{1.6}{2(-0.032)} = 25

and maximum height is

y = 1.6\times 25 - 0.032\times 25^2

y = 20

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards

ratelena [41]3 years ago
5 0

Answer:

<h2>The second ball traveled more distance, horizontally and vertically.</h2>

Explanation:

The given function is

f(x)=-0.032x(x-50)

To find the distances that the second football travels, we need to find the vertex of its movement, because it's movement has a parabola form. The quadratic expression is

f(x)=-0.032x^{2} +1.6x

Where a=-0.032 and b=1.6

The vertex has coordinates of (h,k), where

h=-\frac{b}{2a}

Replacing values, we have

h=-\frac{1.6}{2(-0.032)}=25

Then, k=f(h)

k=f(25)=-0.032(25)^{2} +1.6(25)\\k=-20+40=20

Which means the maximum height of the second football is 20 yards. That means it travels 40 yards vertically.

Now, its horizontal distance can be found when f(x)=0

0=-0.032x^{2} +1.6x\\0=x(-0.032x+1.6)\\x_{1}=0\\ -0.032x_{2} +16=0\\x_{2}=\frac{-16}{-0.032}\\ x_{2}=500

So, its horizontal distance is 500 yards.

Comparing the distances between the footballs.

<h3>Ball 1</h3>

Horizontal distance of 30 yards.

Vertical distance of 30 yards.

<h3>Ball 2</h3>

Horizontal distance of 500 yards.

Vertical distance of 40 yards.

If we find their difference, it would be

Horizontal: 500 - 30 = 470 yards.

Vertical: 40 - 30 = 10 yards.

Therefore, the second ball traveled more distance, horizontally and vertically.

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Answer:

1.34 x 10^3 Pa

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4 0
3 years ago
Find the energy in Joules required to lift a 55.0 Megagram object a distance of 500 cm.
fredd [130]

Energy to lift something =

               (mass of the object) x (gravity) x (height of the lift).

BUT ...

This simple formula only works if you use the right units.

Mass . . . kilograms
Gravity . . . meters/second²
Height . . . meters

For this question . . .

Mass = 55 megagram = 5.5 x 10⁷ grams = 5.5 x 10⁴ kilograms

Gravity (on Earth) = 9.8 m/second²

Height = 500 cm  =  5.0 meters

So we have ...

Energy = (5.5 x 10⁴ kilogram) x (9.8 m/s²) x (5 m)

            =  2,696,925 joules .

That's quite a large amount of energy ... equivalent to
straining at the rate of 1 horsepower for almost exactly an
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7 0
3 years ago
Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m
Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

A=0.0198

From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

d=0.0023

Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

d=0.0120

Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

A=0.0209

From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

Hence, This is the required solution.

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3 years ago
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Answer:

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3 0
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seraphim [82]

The Gay-Lussac's law or Amonton's law states that the pressure of a given amount of a gas is directly propotional to its temperature if its volume is kept constant  .

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and  

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Hence,

Answer is option C

3 0
3 years ago
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