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vodomira [7]
3 years ago
6

Which two formulas are used to calculate potential and kinetic energy

Physics
2 answers:
goblinko [34]3 years ago
6 0

Answer:

gravitational potential energy:

GPE = m g h

kinetic energy:

KE = 1/2 m v^2

kogti [31]3 years ago
3 0

Answer:

\boxed{\bold { \large { \boxed {KE=\frac{1}{2} mv^2 \ , \ PE=mgh}}}}

Explanation:

Kinetic energy formula

\displaystyle KE=\frac{1}{2} mv^2

Potential energy formula

\displaystyle PE=mgh

\displaystyle KE \Rightarrow \sf kinetic \ energy \ (J)

\displaystyle PE \Rightarrow \sf potential \ energy \ (J)

\displaystyle m \Rightarrow \sf mass \ (kg)

\displaystyle v \Rightarrow \sf velocity \ (m/s)

\displaystyle g \Rightarrow \sf acceleration \ of \ gravity\ (m/s^2)

\displaystyle h \Rightarrow \sf height \ (m)

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A wheelchair ramp needs to be 8 inches high how long does it have to be to have a mechanical advantage of 5
Nastasia [14]

MA = (Effort Distance)/(Effort Resistance) = L/H

L = MA * H = 5 * 8" = 40"

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Does lightning McQueen have car or life insurance?
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I am taking an educated guess that he has life insurance
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Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance
PSYCHO15rus [73]

Answer:

No she cannot.

Explanation:

Let v_h be the horizontal component of the ball velocity when it's kicked, assume no air resistance, this is a constant. Also let v_v be the vertical component of the ball velocity, which is affected by gravity after it's kicked.

The time it takes to travel 95m accross the field is

t = 95 / v_h or v_h = 95/t

t is also the time it takes to travel up, and the fall down to the ground, which ultimately stops the motion. So the vertical displacement after time t is 0

s = v_vt + gt^2/2= 0

where g = -9.8m/s2 in the opposite direction with v_v

v_vt - 4.9t^2 = 0

v_vt = 4.9t^2

v_v = 4.9t

Since the total velocity that the goal keeper can give the ball is 30m/s

v = v_v^2 + v_h^2 = 30^2 = 900

(4.9t)^2 + \left(\frac{95}{t})^2 = 900

24.01t^2 + \frac{9025}{t^2} = 900

Let substitute x = t^2 > 0

24.01 x + \frac{9025}{x} = 900

We can multiply both sides by x

24.01 x^2 + 9025 = 900x

24.01x^2 - 900x + 9025 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{900\pm \sqrt{(-900)^2 - 4*(24.01)*(9025)}}{2*(24.01)}

As (-900)^2 - 4*24.01*9025 = -56761 < 0

The solution for this quadratic equation is indefinite

So it's not possible for the goal keeper to do this.

6 0
3 years ago
What is force of gravity​
Eduardwww [97]

Answer:

On Earth all bodies have a weight, or downward force of gravity, proportional to their mass, which Earth's mass exerts on them. Gravity is measured by the acceleration that it gives to freely falling objects. At Earth's surface the acceleration of gravity is about 9.8 metres (32 feet) per second per second.

5 0
3 years ago
A plane is flying east at 115 m/s. The wind accelerates it at 2.88 m/s^2 directly northwest. After 25.0s, what is the magnitude
kondaur [170]

Answer:

81.8 m/s

Explanation:

The initial velocity of the plane is:

v_0=115 m/s (toward east)

So, decomposing along the x- and y- directions:

v_{x0} = 115 m/s\\v_{y0} = 0

(we took east as positive x-direction and north as positive y-direction)

The acceleration is

a=2.88 m/s^2 (northwest, so the angle with the positive x-direction is 135 degrees)

Decomposing it along the two directions:

a_x = a cos 135^{\circ} = (2.88 m/s^2)(cos 135^{\circ})=-2.04 m/s^2\\a_y = a sin 135^{\circ} = (2.88 m/s^2)(sin 135^{\circ})=2.04 m/s^2

So the two components of the velocity after a time t = 25.0 s will be

v_x = v_{x0} + a_x t = 115 m/s + (-2.04 m/s^2)(25.0 s)=64 m/s\\v_y = v_{y0} + a_y t = 0 m/s + (2.04 m/s^2)(25.0 s)=51 m/s

So, the magnitude of the velocity of the plane will be

v=\sqrt[v_x^2+v_y^2}=\sqrt{(64 m/s)^2+(51 m/s)^2}=81.8 m/s

6 0
3 years ago
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