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gulaghasi [49]
3 years ago
11

Amplitude is a Measure of a Waves

Physics
1 answer:
Rasek [7]3 years ago
5 0

C) Energey

A) True

I believe, dont quote me on it

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The greenhouse effect is a result of
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The answer would be D. absorption and radiation of energy in the atmosphere

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If an object measures to be 80 centimeters in length, then it is equivalent to ____​
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31.496 inches

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What is the most effective way to clean the science desks? <br><br> Hypothesis:
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A small sphere of reference-grade iron with a specific heat of 447 J/kg K and a mass of 0.515 kg is suddenly immersed in a water
elena-14-01-66 [18.8K]

Answer:

The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Explanation:

Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer (\dot Q), measured in watts, that is, joules per second, by the following formula:

\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t} (1)

Where:

m - Mass of the sphere, measured in kilograms.

c - Specific heat of the material, measured in joules per kilogram-degree Celsius.

T_{o}, T_{f} - Initial and final temperatures of the sphere, measured in degrees Celsius.

\Delta t - Time, measured in seconds.

In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:

\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}} (2)

Where:

m_{I}, m_{X} - Masses of the iron and unknown spheres, measured in kilograms.

\Delta t_{I}, \Delta t_{X} - Times of the iron and unknown spheres, measured in seconds.

c_{I}, c_{X} - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.

c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}

If we know that \Delta t_{I} = 6.35\,s, \Delta t_{X} = 4.59\,s, m_{I} = 0.515\,kg, m_{X} = 1.263\,kg and c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}, then the specific heat of the unknown material is:

c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)

c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}

Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

3 0
3 years ago
The amount of oxygen in the reactants is 4 atoms. In the products, the oxygen is distributed to water (H2O) and O2 gas. Which co
svp [43]

Answer:

2 in front of water and 1 in front of oxygen

Explanation:

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