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pishuonlain [190]
2 years ago
10

The _____ controls the opening of engine’s valves.

Engineering
1 answer:
mina [271]2 years ago
3 0
A) The Camshaft Controls The Opening of The Engines Valves
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How would an engineer know if a product design were feasible?
Masteriza [31]
- the last one ‘design meets all the criteria...’
8 0
4 years ago
Read 2 more answers
How can you evaluate whether the slope of the dependent variable with an independent variable is the same for each level of the
s2008m [1.1K]

Answer:

By running multiple regression with dummy variables

Explanation:

A dummy variable is a variable that takes on the value 1 or 0. Dummy variables are also called binary

variables. Multiple regression expresses a dependent, or response, variable as a linear

function of two or more independent variables. The slope is the change in the response variable. Therefore, we have to run a multiple regression analysis when the variables are measured in the same measurement.The number of dummy variables you will need to capture a categorical variable

will be one less than the number of categories. When there is no obvious order to the categories or when there are three or more categories and differences between them are not all assumed to be equal, such variables need to be coded as dummy variables for inclusion into a regression model.

3 0
3 years ago
An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at
Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft

G_2 is given as

G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

A=\frac{L}{K}\\A=\frac{460}{115}\\A=4

A is given as

-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%

With initial grade, the elevation of PVC is

E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\

The station is given as

St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\

Low point is given as

x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft

The station of low point is given as

St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\

The elevation is given as

E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

3 0
4 years ago
Show that for a linearly separable dataset, the maximum likelihood solution for the logisitic regression model is obtained by fi
KATRIN_1 [288]

Answer:

Answer for the question:

"Show that for a linearly separable dataset, the maximum likelihood solution for the logisitic regression model is obtained by finding a weight vector w whose decision boundary wx. "

is explained in the attachment.

Explanation:

8 0
3 years ago
Railroad tracks made of 1025 steel are to be laid during the time of year when the temperature averages 4C (40F). Of a joint spa
DENIUS [597]

Answer:

41.5° C

Explanation:

Given data :

1025 steel

Temperature = 4°C

allowed joint space = 5.4 mm

length of rails = 11.9 m

<u>Determine the highest possible temperature </u>

coefficient of thermal expansion ( ∝ ) = 12.1 * 10^-6 /°C

Applying thermal strain ( Δl / l )  = ∝ * ΔT

                                    ( 5.4 * 10^-3 / 11.9 )  = 12.1 * 10^-6 * ( T2 - 4 )

∴  ( T2 - 4 ) =  ( 5.4 * 10^-3 / 11.9 ) / 12.1 * 10^-6

hence : T2 = 41.5°C

8 0
3 years ago
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