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inna [77]
2 years ago
7

A pressure that will support a column of Hg to a height of 256 mm would support a column of water to what height? The density of

mercury is 13.6 g/cm3; the density of water is 1.00 g/cm3.
Physics
1 answer:
Paul [167]2 years ago
6 0

Answer:

<em>The height of water in the column = 348.14 cm</em>

Explanation:

<em>Pressure:</em><em>This is defined as the ratio of the force acting normally ( perpendicular) to the area of surface in contact. The S.I unit of  pressure is N/m²</em>

<em>p = Dgh............... Equation 1</em>

<em>Where p = pressure, D = density, g = acceleration due to gravity, h = height.</em>

<em>From the question, the same pressure will support the column of mercury and water.</em>

<em>p₁ = p₂</em>

<em>Where p₁ = pressure of mercury, p₂ = pressure of water</em>

D₁gh₁ = D₂gh₂.................. Equation 2

making h₂ the subject of equation 2

h₂ = D₁gh/D₂g............... Equation 3

Where D₁ and D₂ = Density of mercury and water respectively, h₁ and h₂ = height of mercury and water respectively

Given: D₁ = 13.6 g/cm³, D₂ = 1.00 g/cm³, h₁ = 256 mm = 25.6 cm.

Constant: g = 9.8 m/s²

Substituting these values into Equation 3,

h₂ = (13.6×9.8×25.6)/1×9.8

<em>h₂ = 348.14 cm</em>

<em>The height of water in the column = 348.14 cm</em>

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Answer:

a) Null hypothesis:  \mu \geq 16  

Alternative hypothesis :\mu  

b) t=\frac{15.6-16}{\frac{0.3}{\sqrt{8}}}=-3.77  

p_v =P(t_{(7)}  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis.

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Explanation:

Data given and notation  

\bar X=15.6 represent the mean for the sample

s=0.3 represent the sample standard deviation  

n=8 sample size  

\mu_o =16 represent the value that we want to test  

\alpha=0.1 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

Is a one tailed left test.  

What are H0 and Ha for this study?  

Null hypothesis:  \mu \geq 16  

Alternative hypothesis :\mu  

The degrees of freedom on this case are:

df=n-1=8-1=7

Compute the test statistic

The statistic for this case is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{15.6-16}{\frac{0.3}{\sqrt{8}}}=-3.77  

Give the appropriate conclusion for the test

Since is a one side left tailed test the p value would be:  

p_v =P(t_{(7)}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis.

We can conclude that the mean is significantly less than 16 ounces at 10% of significance.  so then the consumer suspect is correct.

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