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3 years ago

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You have purchased an inexpensive USB oscilloscope (which measures and displays voltage waveforms). You wish to determine if the

oscilloscope has an error bias; in other words, you wish to determine if the errors made by the oscilloscope have a population mean that is not equal to zero. So you use a very accurate voltmeter to find the measurement errors for 13 different measurements made by your USB oscilloscope. A data file containing these measurements is HTMean1.csvPreview the document . Do a statistical analysis on this data to determine if the oscilloscope has an error bias.

1 answer:

4vir4ik [10]3 years ago

7 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

1

A

2

A

Explanation:

From the question the data given for Error (mV) is -15

-15.17

8.67

-13.74

-20.69

-6.96

-1.36

-2.96

-9.26

3.11

-14.12

6.39

-14.77

Generally

The null hypothesis is $H_o : \mu = 0$

The alternative hypothesis is $H_a : \mu \ne 0$

The sample size is n = 13

Here $\mu$ represents the true error bias (i.e population error bias)

Generally the sample error bias is mathematically represented as

$\= x = \frac{ \sum x_i}{n}$

=> $\= x = \frac{ -15.17 + 8.67 + (-13.74) + \cdots + (-14.77) }{13}$

$\= x = -7.37$

Generally the standard deviation is mathematically represented as

$\sigma = \sqrt{\frac{\sum (x_i - \= x )^2}{n} }$

=> $\sigma = \sqrt{\frac{ (-15.17-( -7.37) )^2 + (8.67 -( -7.37) )^2 + \cdots + (-14.77 -( -7.37) )^2 }{13} }$

=> $\sigma = \sqrt{ 119.385}$

=> $\sigma = 10.926$

Generally the test statistics is mathematically represented as

$t = \frac{\= x - \mu }{\frac{\sigma }{\sqrt{n} } }$

=> $t = \frac{ -7.37 - 0 }{\frac{10.926}{\sqrt{13} } }$

=> $t = -2.838$

Generally the p-value is mathematically represented as

$p-value = 2 P(t < -2.432)$

From the z-table $P(t < -2.432) = 0.0075$

So $p-value = 2* 0.0075$

=> $p-value = 0.015$

So given that p-value is less than the $\alpha = 0.05$ then we reject the null hypothesis and conclude that the oscilloscope has an error bias

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Answer:

The neutron loses all of its kinetic energy to nucleus.

Explanation:

Given:

Mass of neutron is 'm' and mass of nucleus is 'm'.

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Now, final kinetic energy of the system will be equal to the initial kinetic energy.

Now, as the nucleus was at rest initially, so the final kinetic energy of the nucleus will be equal to the initial kinetic energy of the neutron.

Thus, all the kinetic energy of the neutron will be transferred to the nucleus and the neutron will come to rest after collision.

Therefore, the neutron loses all of its kinetic energy to nucleus.

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3 years ago

efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.

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Answer:

Inlet : $v_i=0.0646\frac{m}{s}$

Outlet: $v_o=0.171\frac{m}{s}$

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

$\dot{m}=\frac{\dot{V}}{\upsilon}$ , where $\dot{V}$ represent the flow rate and $\upsilon$ the specific volume at the pressure and temperature given.

$A_i=0.5m^2$ is the inlet area

$P_i=600Kpa$ pressure at the inlet area

$T_i=70C$ temperature at the inlet area

$A_o=1m^2$ is the outlet area

$P_o=100Kpa$ pressure at the outlet area

$T_o=C$ temperature at the outlet area

$\dot{m}=0.75\frac{kg}{s}$ represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

$\upsilon_i =0.04304\frac{kg}{m^3}$ and the entropy is $h_i=1.0645\frac{KJ}{KgK}=h_o$

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy $h_o=1.0645\frac{KJ}{KgK}$

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

$h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}$

$h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}$

Our interest value would be given using interpolation like this:

$\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}$

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that $\dot{m}=\frac{\dot{V}}{\upsilon}$ , but since $\dot{V}=Av$ we have this:

$\dot{m}=\frac{Av}{\upsilon}$

If we solve from the velocity v we have this:

$v=\frac{\upsilon \dot{m}}{A}$ (*)

And now we just need to replace the values into equation (*)

For the inlet case:

$v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}$

For the oulet case:

$v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}$

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3 years ago

A spy satellite uses a telescope with a 1.7-m-diameter mirror. It orbits the earth at a height of 180 km.

WINSTONCH [101]

Answer: the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm

Explanation:

Given that;

diameter of the mirror d = 1.7 m

height h = 180 km = 180 × 10³ m

wavelength λ = 500 nm = 5 × 10⁻⁹ m

Now Angular separation from the peak of the central maximum is expressed as;

sin∅= 1.22 λ / d

sin∅ = (1.22 × 5 × 10⁻⁹) / 1.7

sin∅ = 3.588 × 10⁻⁷

we know that;

sin∅ = object separation / distance from telescope

object separation = sin∅ × distance from telescope

object separation = 3.588 × 10⁻⁷ × 180 × 10³

object separation =6.45 × 10⁻² m

then we convert to centimeter

object separation = 6.45 cm

Therefore the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm

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3 years ago

A beam of light, initially travelling in the air, strikes water surface at an angle of 24.5° with the normal. If the speed of li

kykrilka [37]

Answer:

Explanation:

n = 3.00e8/2.22e8 = 1.35

1.00sin24.5 = 1.35sinθ

θ = 17.9°

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2 years ago

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Answer:

The answer to the question is;

The total potential energy of the mass on the spring when the mass is at either endpoint of its motion is 5.0255 Joules.

Explanation:

To answer the question, we note that the maximum speed is 2.30 m/s and the mass is 1.90 kg

Therefore the maximum kinetic energy of motion is given by

Kinetic Energy, KE = $\frac{1}{2} mv^{2}$

Where,

m = Attached vibrating mass = 1.90 kg

v = velocity of the string = 2.3 m/s

Therefore Kinetic Energy, KE = $\frac{1}{2}$ ×1.9×2.3² = 5.0255 J

From the law of conservation of energy, we have the kinetic energy, during the cause of the vibration is converted to potential energy when the mass is at either endpoint of its motion

Therefore Potential Energy PE at end point = Kinetic Energy, KE at the middle of the motion

That is the total potential energy of the mass on the spring when the mass is at either endpoint of its motion is equal to the maximum kinetic energy.

Total PE = Maximum KE = 5.0255 J.

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