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Temka [501]
3 years ago
10

What is the difference between a frequency polygon and an​ ogive? Choose the correct answer below.

Physics
1 answer:
larisa86 [58]3 years ago
6 0
The difference between the two is that frequency polygon is a graph that presents the information by using lines that connect points plotted for the frequencies at the midpoints of the classes. The frequencies are signified by the heights of the points while the the ogive is a graph that signifies the cumulative frequencies for the classes in a frequency distribution.
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Which would melt first, germanium with a melting point of 1210 k or gold with a melting point of 1064oc?
iren2701 [21]
<span>Germanium To determine which melts first, convert their melting temperatures so they're both expressed on same scale. It doesn't matter what scale you use, Kelvin, Celsius, of Fahrenheit. Just as long as it's the same scale for everything. Since we already have one substance expressed in Kelvin and since it's easy to convert from Celsius to Kelvin, I'll use Kelvin. So convert the melting point from Celsius to Kelvin for Gold by adding 273.15 1064 + 273.15 = 1337.15 K So Germanium melts at 1210K and Gold melts at 1337.15K. Germanium has the lower melting point, so it melts first.</span>
8 0
3 years ago
What is it that makes a magnet different from a piece of iron that is not mangetic?​
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Answer:

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7 0
3 years ago
) Water flows through a horizontal coil heated from the outside by high-temperature flue gases. As it passes through the coil th
Mademuasel [1]

Explanation:

Formula for steady flow energy equation for the flow of fluid is as follows.

    m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w

Now, we will substitute 0 for both z_{1} and z_{2}, 0 for w, 334.9 kJ/kg for h_{1}, 2726.5 kJ/kg for h_{2}, 5 m/s for V_{1} and 220 m/s for V_{2}.

Putting the given values into the above formula as follows.

     m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w  

     1 \times [334.9 \times 10^{3} J/kg + \frac{(5 m/s)^{2}}{2} + 0] + q = 1 \times [2726.5 \times 10^{3} + \frac{(220 m/s)^{2}}{2} + 0] + 0

                q = 6597.711 kJ

Thus, we can conclude that heat transferred through the coil per unit mass of water is 6597.711 kJ.

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