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elena-s [515]
2 years ago
9

A 20-turn coil of area 0.32 m2 is placed in a uniform magnetic field of 0.055 T so that the perpendicular to the plane of the co

il makes an angle of 30∘ with respect to the magnetic field.
The flux through the coil is
Physics
1 answer:
makvit [3.9K]2 years ago
5 0

Answer:

1.5 * 10^-2 Tm^2

Explanation:

Electric Flux = B.A cos(theta)

B = 0.055 T

A = 0.32 m^2

theta = 30

Electric Flux = (0.055 T).(0.32 m^2).Cos(30) = 0.0152 = 1.5 * 10^-2 Tm^2

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While traveling along a highway a driver slows from 31 m/s to 15 m/s in 8 seconds. What is the automobile’s acceleration? (Remem
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Answer:

The automobile's acceleration in that time interval is -2 m/s^2

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In this case, we have:

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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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