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Kazeer [188]
3 years ago
15

Arrange the examples in order, starting with the object that has the least amount of energy. In each case, assume there’s no fri

ction. Use g = 9.8 m/s2, PE = m × g × h, and . Tiles a book with a mass of 0.75 kilograms resting on a shelf at a height of 1.5 metersa brick with a mass of 2.5 kilograms falling with a velocity of 10 meters/second when it’s 4 meters above grounda ball with a mass of 0.25 kilograms rolling on flat ground with a velocity of 10 meters/seconda stone with a mass of 0.7 kilograms being held still at a height of 7 meters
Physics
1 answer:
Artemon [7]3 years ago
4 0
First example: book, m= 0.75 kg, h=1.5 m, g= 9.8 m/s², it has only potential energy Ep,

Ep=m*g*h=0.75*9.8*1.5=11.025 J

Second example: brick, m=2.5 kg, v=10 m/s, h=4 m, it has potential energy Ep and kinetic energy Ek,

E=Ep+Ek=m*g*h + (1/2)*m*v²=98 J + 125 J= 223 J

Third example: ball, m=0.25 kg, v= 10 m/s, it has only kinetic energy Ek

Ek=(1/2)*m*v²=12.5 J.

Fourth example: stone, m=0.7 kg, h=7 m, it has only potential energy Ep,

Ep=m*g*h=0.7*9.8*7=48.02 J

The order of examples starting with the lowest energy:

1. book, 2. ball, 3. stone, 4. brick 


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<span>1) The differential equation that models the RC circuit is :

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2)
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The solution of the differential equation is

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And the maximun Voltage the capacitor will have in this configuration is the voltage of the battery  V_capac(∞) = 9V 

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V_capac (t) = (-9V)e ^(-t /T)  +  9V

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Where C is the capacitance and V is the Voltage across

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /T)  +  9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s)  +  9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s)  +  9V = -1.87V +9V

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Q (4.3048 s)  = 89.9mF*(7.1275V) = 0.6407 C

3) The charge after a very long time refers to the maximum charge the capacitor will hold in this circuit. This occurs when the voltage accross its terminals is equal to the voltage of the battery = 9V

Q (∞)  = 89.9mF*(9V) = 0.8091 C
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