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3 years ago

Compare and contrast the average kinetic energy of 0.5 L of coffee at 34ÁC,

1 answer:

6 0

Compared to coffee at room temperature, the molecules of the coffee at 34°C will be moving faster and colliding with one another more frequently.

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temperature, atmosphere pressure,wind, topography, humidity etc

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3 years ago

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2 years ago

A coffee filter of mass 1.2 grams dropped from a height of 1 m reaches the ground with a speed of 0.8 m/s. How much kinetic ener

iren [92.7K]

Answer: The energy gained by the air molecules is 0.011 J.

Explanation:

Law of conservation of energy states that energy can neither be created nor be destroyed but it can only be transformed from one form to another form.

Here, the potential energy of the coffee filter is getting converted into kinetic energy of the coffee filter and some energy is lost by it which is gained by the air molecules in the form of kinetic energy.

So, calculating the potential energy of coffee filter, we use the equation:

P = mgh

where,

m = mass of coffee filter = 1.2 g = $1.2\times 10^{-3}kg$ (Conversion used: 1 kg = 1000 g)

g = acceleration due to gravity = $9.8m/s^2$

h = height of coffee filter = 1 m

Putting values in above equation, we get:

$P=1.2\times 10^{-3}kg\times 9.8m/s^2\times 1m\\\\P=1.176\times 10^{-2}J$

Calculating the kinetic energy of coffee filter, we use the equation:

$E=\frac{1}{2}mv^2$

where,

m = mass of coffee filter = 1.2 g = $1.2\times 10^{-3}kg$

v = speed of coffee filter = 0.8 m/s

Putting values in above equation, we get:

$E=\frac{1}{2}\times 1.2\times 10^{-3}kg\times (0.8m/s)^2\\\\E=3.84\times 10^{-4}J$

As, energy lost by coffee filter = energy gained by air molecules

So, energy lost by coffee filter = Potential energy - Kinetic energy

Energy lost by coffee filter = $(1.176\times 10^{-2})-(3.84\times 10^{-4})=0.011J$

Hence, the energy gained by the air molecules is 0.011 J.

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3 years ago

Una locomotiva traina un solo vagone, che ha la sua stessa massa, con la forza di

gregori [183]

Answer:

c) F = 16000 N

Explanation:

For this exercise we use Newton's second law

F = ma

they tell us that adding the other wagons the acceleration of the locomotive must be maintained

F = m a

by adding the other four wagons

mass = 4 no

therefore to maintain the force you must also raise the same factor

Fe = 4Fo

Fe = 4 4000

F = 16000 N

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3 years ago