2 years ago

When hitting the ski slopes, which location do skiers have the most potential energy

Mathematics

1 answer:

lisabon 2012 [21]2 years ago

5 0

Answer:

at the top of the mountain or hill

Step-by-step explanation:

potential energy is stored energy so therefore the higjestpoint will have the most potential energy.

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F(x)=x^2+3x-2 find f(-2)

Rasek [7]

Answer:

-12

Step-by-step explanation:

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2 years ago

The hanger image below represents a balanced equation.

Allushta [10]

Answer:

2x =3

The hanger image given in the figure represents balanced equation.

Then we have to write an equation to represent the image.

In balanced form left side = right side

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1 year ago

PLEASE HELP

Aliun [14]

Hello from MrBillDoesMath!

Answer:

(x^4-2) (x^4 -1)

Discussion:

Let u = x^4, then

x^8 - 3x^4 + 2

= u^2 - 3u + 2

= (u -2) ( u -1)

= (x^4-2) (x^4 -1)

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3 years ago

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Alexeev081 [22]

My numbers aren't exactly the same due to rounding but you can still see the answer.

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2 years ago

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In this exercise, consider a particle moving on a circular path of radius b described by r(t) = b cos(ωt)i + b sin(ωt)j, where ω

Bingel [31]

Answer:

Acceleration of the particle = $bw^{2}$

Step-by-step explanation:

We are given the position vector of a particle moving in a circle of radius b units.

r(t) = b cos(ωt)i + b sin(ωt)j

Velocity , v = $\frac{dr}{dt}$ = -bω sin(ωt)i + bω cos(ωt)j

The magnitude of velocity, v = $\sqrt{v_x^{2} +v_y^{2} }$

Squaring both sides,

$v^{2} = b^{2} w^{2}(sin^{2}(wt)+cos^{2}(wt))$

Since $sin^{2}(wt)+cos^{2}(wt))$ = 1

$v^{2} = b^{2}w^{2}$

The acceleration towards the centre is called the centripetal acceleration and is given by

a = $\frac{v^{2} }{r}$

a = $\frac{b^{2}w^{2}}{b}$

a = $bw^{2}$

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2 years ago