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Sophie [7]
4 years ago
6

Mr. Johnson challenged his students to see which group could dissolve salt the fastest in 100 ml of water. The students could ch

ange any variable(s) except the volume of water used. Group 1 changed the water temperature. They used tap water at 22oC and then three other temperatures: 32, 42, and 52 oC. They did not stir the mixture and they kept the volume of the water constant, just like Mr.Johnson directed. The salt dissolved fastest in the warmest water. In their conclusions, Group 1 wanted to include a line graph of their data. 
What label(s) should they place on the X axis?

A)rate of dissolving. B)volume of water used. C)time to dissolve (sec). D)temperature of water
Physics
2 answers:
Svetradugi [14.3K]4 years ago
7 0

Answer:

D)temperature of water

Explanation:

To understand why the "water temperature" would be plotted on the X axis, we must first understand what the dependent variables and the independent variables are.

Dependent variables are those that are influenced by another variable to happen. This type of variable can be measured and is usually the basis of the result of the experiment. On the other hand, the independent variable is the one that influences the dependent variable to show a result. In the case shown in the question above, the independent variable is the water temperature, because it influences the dissolution rate (dependent variable) of the salt.

In the graphical representation of the association between two variables, the independent variable is plotted on the X axis. The dependent variable is plotted on the Y axis.

babunello [35]4 years ago
6 0
I'm not sure its correct but the answer could be, C)time to dissolve (sec).
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At noon, ship A is 140 km west of ship B. Ship A is sailing east at 30 km/h and ship B is sailing north at 25 km/h. How fast (in
Luba_88 [7]

Answer:

The rate of change of distance between the two ships is 18.63 km/h

Explanation:

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2r(\frac{dr}{dt} ) = 2a(\frac{da}{dt} )  + 2b(\frac{db}{dt} )\\\\r(\frac{dr}{dt} ) = a(\frac{da}{dt} )  + b(\frac{db}{dt} )\\\\101.98(\frac{dr}{dt} ) = 20(-30 )  + 100(25 )\\\\101.98(\frac{dr}{dt} ) = -600 + 2,500\\\\101.98(\frac{dr}{dt} ) = 1900\\\\\frac{dr}{dt}  = \frac{1900}{101.98} = 18.63 \ km/h

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