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aniked [119]
3 years ago
6

The complete combustion of a small wooden match produces approximately 512 cal of heat. How many kilojoules are produced? Expres

s your answer in terms of kilojoules.
Physics
1 answer:
k0ka [10]3 years ago
5 0

We have to covert 512 cal of heat in kilo joules.

As, 1 cal = 0.004184 kJ = 4.184 joules.

Therefore,

512 \ cal = 512 \times 4.184 \ J = 2142.208 \ J \\\\\ =  2.142 \times 10 ^3 J = 2.142 \ kJ

Thus, combustion of a small wooden match produces approximately ( in kilo joules ) is 2.142 kJ .

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A 10-μF capacitor in an LC circuit made entirely of superconducting materials ( R = 0 Ω ) is charged to 100 μC. Then a supercond
satela [25.4K]

Answer:

Vdc=10V

Explanation:

in a closed loop consisting of a super charged capacitor and an inductor, the super charge capacitor acts as a supply when the loop is closed, at t=0, the emf stored in the capacitor is 10V (q/c); and at that same time Vl= voltage across the inductor or loop too would be 10V,

if the loop remains closed for  a longer period, the inductor would absorb energy from the capacitor till it dissipates all charges with itself.

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1. What is the arrangement of the outer planets? 2. What effect does their placement have the planets?
Rasek [7]
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<span>4. Mars </span>
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Give an example of where you see the law of conservation of energy in your life
lesya692 [45]

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The law of conservation of energy can be seen in these everyday examples of energy transference: Water can produce electricity. Water falls from the sky, converting potential energy to kinetic energy.

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Explanation:

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What are the four system of measurement ​
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ask google

Explanation:

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4 years ago
Se lanza verticalmente una esfera con una rapidez de 30m/se. Determinar la rapidez de la esfera a una altura de 40m (g=10m/s2)
sergeinik [125]

v^2-{v_0}^2=2a(x-x_0)

dónde v es la velocidad de la esfera, v_0 es suya velocidad inicial, a=-g la aceleración debida a la gravedad, x la posición, y x_0 la posición inicial. Tomamos x_0=0\,\mathrm m a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

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\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}

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