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ExtremeBDS [4]
3 years ago
9

What is the gravitational potential energy of a 0.1 kg apple sitting on top of a 12.5 m flagpole? Note: GPE= mgh. m= mass, g= ac

celeration due to gravity (9.8 m/s^2),
h= objects height above the ground.
Physics
1 answer:
Pachacha [2.7K]3 years ago
6 0
<span>GPE= mgh = 0.1 kg * 9.8 m/s^2  * 12.5m = 12.25 J</span>
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A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim
Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

F_D = -\frac{1}{2}\rho V^2 C_d A

Where

\rho is the density of the flow

V = Velocity

C_d= Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

F=ma=m\frac{dV}{dt}

Equating both equations we have:

m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A

m(dV)=-\frac{1}{2}\rho C_d A (dt)

\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)

Integrating

\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)

-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t

Here,

V_f = 60mph = 26.82m/s

V_i = 120.7m/s

m= 1600lbf = 725.747Kg

\rho = 1.21 kg/m^3

C_d = 0.3

\Delta t=7s

Replacing:

\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)

-0.029 = -5.4997r^2

r = 2.2963m

d= r*2 = 4.592m \approx 15.065ft

4 0
3 years ago
Roughly to what height would a 5 kg stone need to be raised in order to have the same stored energy as the energy stored in the
hoa [83]

8.16m is the required height, a 5kg stone need to be raised.

One sort of potential energy is gravitational potential energy, which is equal to the product of the object's mass (m), the gravitational acceleration (g), and the object's height (h) as measured in relation to the ground's surface (the body).

We obtain the formula by considering the work done in raising a mass m through a height h.

Work in elevating mass m through height h is equal to force times distance.

The force must be greater than the mass m's weight, hence F = mg.

Work done = mgh = gravitational potential energy

Energy =  Mass of the object × gravitational acceleration × height.

Mass of the stone = 5kg

Equating ;

∴ 400 J = 5 kg × 9.8 m/s² × height

  Height = 8.16 m

Therefore, 8.16m is the required height.

Learn more about energy here:

brainly.com/question/1242059

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Answer:

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Identify two size dependent properties and two size independent properties of an iron nail
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hope this helps

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Velocity must be measured in units of
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