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3 years ago

Woman pulls a 6.87 kg suitcase,

Physics

1 answer:

GenaCL600 [577]3 years ago

4 0

Answer:

2.24 m/s

Explanation:

resolving force of 29.2 N in x component

Fx = 29.2 cos 57.7

Fx = 15.6N

as force of friction is 12.7 N hence net force which produces acceleration is

15.6-12.7=2.9 N

by Newton 's law a=f/m

a= 2.9/6.87=0.422 m/s^2

now equation of motion is

v^2= U^2+2as

= 0^2+2(.422)(5.93)

v^2=5.00

v=2.24 m/s

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3 years ago

If you were given distance and period of time, what can you calculate?

antiseptic1488 [7]

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3 years ago

When an object in simple harmonic motion is at its maximum displacement, its____________ is also at a maximum.

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When an object in simple harmonic motion is at its maximum displacement, its acceleration is also at a maximum.

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2 years ago

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

babunello [35]

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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3 years ago

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blondinia [14]

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