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guajiro [1.7K]
3 years ago
12

The diagram below represents a flashlight that has been turned on. Which form of energy is being converted to electrical energy

by the batteries in the flashlight?
Physics
1 answer:
Orlov [11]3 years ago
3 0

Answer:

Chemical energy to electrical energy

Explanation:

In nature, there are several types of energy.

In this example (a flashlight being turned on), we have a conversion of energy from chemical energy to electrical energy. In fact:

- Chemical energy is the energy stored in the chemical bonds of the molecules of the substances used inside the battery. When the chemical reaction inside the battery occurs, this energy is liberated, and it is used to "push" the electrons along the circuit connected to the battery

- Electric energy is the energy associated to the motion of the electrons along the circuit of the flashlight; it is the energy associated to an electric current.

Moreover, in the flashlight the electric energy is then converted into two more types of energy: light energy (since the bulb in the flashlight produces light) and heat energy (because the flashlight also produces heat, so thermal energy).

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A sound wave travelling in water has a frequency of 100 Hz. The speed of sound in water is 1482 m/s. Calculate the wavelength of
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λ = v/f
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The nucleus of most atoms is composed of which of the following sub-atomic particles?
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Helium-oxygen mixtures are used by divers to avoid the bends and are used in medicine to treat some respiratory ailments. What p
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Answer:

The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%

Explanation:

From General gas equation.

PV = nRT...............................  Equation 1

Where n = number of moles, V = volume, P = pressure, T = temperature, P = pressure, V = volume.

n = mass/molar mass .................. Equation 2

substituting equation 2 into equation 1.

PV = (mass/molar mass)RT

⇒ Mass/molar mass = PV/RT..................... Equation 3

But mass = Density × Volume

⇒ M = D × V.................... Equation 4

Where D = density, M = mass

Substituting equation 4 into equation 3

DV/molar mass = PV/RT............ Equation 5

Dividing both side of the equation by Volume (V) in Equation 5

D/molar mass = P/RT .............. Equation 6

Cross multiplying equation 6

D × RT = P × molar mass

∴ Molar mass = (D × RT)/P.................. Equation 7

Where D = 0.518 g/L , R = 0.0821 atm dm³/K.mol,

T = 25°C = 25 + 273 = 298 K,

P =721 mmHg = (721/760) atm= 0.949 atm

Substituting these values into equation 7

Molar mass = (0.518 × 0.0821 × 298)/0.949

Molar mass = 13.35 g/mole

The molar mass of the mixture is =13.35 g/mole

Let y be the mole fraction of Helium and 1-y be the mole fraction of oxygen.

∴ 13.35 = 4(y) + 32(1-y)

13.35 = 4y + 32 - 32y

Collecting like terms in the equation,

32y - 4y = 32 - 13.35

28y = 18.65

y = 18.65/28

y =0.666

y = 0.666 × 100 = 66.6%

∴The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%

6 0
3 years ago
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