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lukranit [14]
3 years ago
9

Jane is collecting data for a ball rolling down a hill. she measure out a set of different distances and then proceeds to use a

stopwatch to find the time it takes the ball to roll each distance
Physics
1 answer:
Snezhnost [94]3 years ago
5 0

Answer:

<em>The Independent variable in this experiment is the time taken by the ball to roll down each distance.</em>

<em>The dependent variable is the distance  through which the ball rolls</em>

<em>The control variables are: slope of hill, weight, of the ball, size of ball, wind speed, surface characteristics of the ball.</em>

Explanation:

The complete question is

Jane is collecting data for a ball rolling down a hill. She measures out a set of different distances and then proceeds to use a stop watch to find the time it takes the ball to roll. What are the independent, dependent, and control variables in this experiment?

Independent variable have their values not dependent on any other variable in the scope of the experiment. The time for the ball to roll down the hill is not dependent on any other variable in the experiment. Naturally, <em>some common independent variables are time, space, density, mass, fluid flow rate.</em>

A dependent variable has its value dependent on the independent variable in the experiment. The value of the distance the ball rolls depends on the time it takes to roll down the hill.

The relationship between the dependent and independent variables in an experiment is given as

y = f(x)

where y is the output or the dependent variable,

and x is the independent variable.

Control variables are those variable that if not held constant could greatly affect the results of an experiment. For an experiment to be more accurate, control variables should be confined to a given set of value throughout the experiment.

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Answer:

P_{sol}=50.4\ mm.Hg

Explanation:

According to given:

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<em>Upon the formation of solution the vapour pressure will be reduced since we have one component of solution as non-volatile.</em>

<u>moles of water in the given quantity:</u>

n_w=\frac{m_w}{M_w}

n_w=\frac{313.5}{18}

n_w=17.42 moles

<u>moles of glycerin in the given quantity:</u>

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n_g=\frac{154}{92}

n_g=1.674 moles

<u>Now the mole fraction of water:</u>

X_w=\frac{n_w}{n_w+n_g}

X_w=\frac{17.42}{17.42+1.674}

X_w=0.912

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\therefore P_{sol}=X_w\times P_x

\therefore P_{sol}=0.912\times 55.32

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(way 2) using conservation of energy
energy (KE+PE) on top = energy at bottom
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