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3 years ago

7

The blades of a fan running at low speed turn at 26.2 rad/s. When the fan is switched to high speed, the rotation rate increases

uniformly to 36.5 rad/s in 5.75 seconds. What is the magnitude of the fan's angular acceleration

2 answers:

Lady bird [3.3K]3 years ago

8 0

Answer: $1.79\ rad/s^2$

Explanation:

Given

Initial angular speed is $\omega_1=26.2\ rad/s$

Final angular speed is $\omega_2=36.5\ rad/s$

Time period $t=5.75\ s$

Magnitude of the fan's acceleration is given by

$\Rightarrow \alpha=\dfrac{\omega_2-\omega_1}{t}$

Insert the values

$\Rightarrow \alpha=\dfrac{36.5-26.2}{5.75}\\\\\Rightarrow \alpha=\dfrac{10.3}{5.75}\\\\\Rightarrow \alpha=1.79\ rad/s^2$

Thus, fan angular acceleration is $1.79\ rad/s^2$

valina [46]3 years ago

8 0

Answer:

The angular acceleration is given by 1.8 rad/s^2.

Explanation:

initial angular speed, wo = 26.2 rad/s

final angular velocity, w = 36.5 rad/s

time, t = 5.75 seconds

The first equation of motion is

$w = wo + \alpha t\\\\36.5 = 26.2 + 5.75\alpha\\\\\alpha = 1.8 rad/s^2$

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A machine is designed to fill jars with 16 ounces of coffee. A consumer suspects that the machine is not filling the jars comple

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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.

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

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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