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2 years ago

What is the step in which you are testing your hypothesis

Engineering

1 answer:

jonny [76]2 years ago

8 0

Answer:

Step 1: State your null and alternate hypothesis. ...

Step 2: Collect data. ...

Step 3: Perform a statistical test. ...

Step 4: Decide whether the null hypothesis is supported or refuted. ...

Step 5: Present your findings.

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8. Describe and correct the error in stating the domain. Xf * (x) = 4x ^ (1/2) + 2 and g(x) = - 4x ^ (1/2) The domain of (f + g)

konstantin123 [22]

Answer:

Explanation:

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2 years ago

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sergejj [24]

Answer:

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8 0

2 years ago

The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature

Anit [1.1K]

Answer:

a). Work transfer = 527.2 kJ

b). Heat Transfer = 197.7 kJ

Explanation:

Given:

$P_{1}$ = 5 Mpa

$T_{1}$ = 1623°C

= 1896 K

$V_{1}$ = 0.05 $m^{3}$

Also given $\frac{V_{2}}{V_{1}} = 20$

Therefore, $V_{2}$ = 1 $m^{3}$

R = 0.27 kJ / kg-K

$C_{V}$ = 0.8 kJ / kg-K

Also given : $P_{1}V_{1}^{1.25}=C$

Therefore, $P_{1}V_{1}^{1.25}$ = $P_{2}V_{2}^{1.25}$

$5\times 0.05^{1.25}=P_{2}\times 1^{1.25}$

$P_{2}$ = 0.1182 MPa

a). Work transfer, δW = $\frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}$

$\left [\frac{5\times 0.05-0.1182\times 1}{1.25-1} \right ]\times 10^{6}$

= 527200 J

= 527.200 kJ

b). From 1st law of thermodynamics,

Heat transfer, δQ = ΔU+δW

= $\frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}$

= $\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W$

= $\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200$

= 197.7 kJ

6 0

2 years ago

One half of 17’ 9 3/16”

Mademuasel [1]

555.66 is the answer I think

6 0

2 years ago

120 litres of water is discharge from container in 25 seconds. Find the rate of discharge in cumecs.if the discharge took place

notsponge [240]

<h2>Answer:</h2>

Rate of discharge in cumecs: <u>0.0048m³/s</u>.

Velocity flow: <u>24m/s</u>.

<h2>Explanation:</h2>

<h3>1. Find the rate of discharge in cumecs.</h3>

a. Convert from litres to m³.
120L*1000= 120000mL

120000mL=120000cm³

120000cm³/100³=0.12 m³.

b. Rate of discharge.

<em>If 0.12 m³ where discharged in 25 seconds, the rate of discharge is:</em>

0.12m³/25s = 0.0048m³/s.

<em />

<h3>2. Find the velocity flow.</h3>

Let's refer to the fluid mechanics equation that relates volume flow, area and velocity. This is the formula:

$\frac{dV}{dt}=Av$ ; where the expression $\frac{dV}{dt}$ is the volume flow rate (in m³/s); A is the cross-sectional area of the pipe (in m²), and v is the velocity flow (in m/s).

a. Solve the equation for v.

$\frac{dV}{dt}=Av\\ \\(\frac{dV}{dt})/A=v\\ \\v=(\frac{dV}{dt})/A$

b. Calculate the cross-sectional area of the pipe.

<em>The cross-sectional area of the pipe is a circle. Hence, the formula of this area is:</em>

$A=\pi r^{2}$

<em>We'll have to convert the diameter to meters, because the formula for flow velocity needs the area in m². Let's go ahead and do that.</em>

<em /> $50mm/1000=0.05m$ .

<em>We were given the diameter, and the formula uses the radius, but the radius is just half of the diameter, therefore, we can substitute in toe formula like this:</em>

$A=\pi (\frac{0.05}{2} )^{2}=0.0020m^{2}$

c. Substitute in the new expression for velocity flow and calculate.

$v=(\frac{dV}{dt})/A\\ \\v=(\frac{0.048m^{3} }{1s})/(0.0020m^{2} )\\\\ v= 24m/s$

8 0

1 year ago

What is the step in which you are testing your hypothesis ​

What is the step in which you are testing your hypothesis