Natural-draft furnaces are no longer available for new installations because they are ____

What material property would still cause strain in a strain gauge that is positionedperpendicular to the direction of force if i

svetlana [45]

Answer:

oof

Explanation:

I don't know but please don't report me

I am trying to do a challenge

Thank you-

If you don't report me!

5 0

2 years ago

Write a SELECT statement that returns the same result set as this SELECT statement. Substitute a subquery in a WHERE clause for

sergiy2304 [10]

Answer:

SELECT distinct VendorName FROM Vendors

WHERE VendorID IN (

SELECT VendorID FROM Invoices

)

Explanation:

6 0

3 years ago

Analyzing Manufacturing Production Process Development Tasks

artcher [175]

Answer:

4, 5, 6

Explanation:

i just did it

5 0

3 years ago

Read 2 more answers

The mean of hours that the average person watches television each day is 4.18 hours with a standard deviation of 1.19 hours. Fin

luda_lava [24]

Answer:

$z = \frac{3-4.18}{1.19}=-0.992$

$z = \frac{5-4.18}{1.19}=0.689$

And we can find this probability with this difference:

$P(-0.992$

And then we can conclude that the probability that someone watches between 3 and 5 hours a day is approximately 0.591 using a normal distribution

Explanation:

For this case we can define the random variable X as "hours that a person watches television". For this case we don't have the distribution for X but we have the following parameters:

$\mu = 4.18,\sigma =1.19$

We can assume that the distribution for X is normal

$X \sim N(\mu = 4.18 , \sigma =1.19)$

And we want to find this probability:

$P(3$

And we can use the z score formula given by:

$z=\frac[X- \mu}{\sigma}$

And we can find the z score for each limit and we got:

$z = \frac{3-4.18}{1.19}=-0.992$

$z = \frac{5-4.18}{1.19}=0.689$

And we can find this probability with this difference:

$P(-0.992$

And then we can conclude that the probability that someone watches between 3 and 5 hours a day is approximately 0.591 using a normal distribution

6 0

2 years ago

A rectangular car-top carrier of 1.7-ft height, 5.0-ft length (front to back), and 4.2-ft width is attached to the top of a car.

Nataliya [291]

Answer:

$\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp$

Explanation:

We can assume that the general formula for the drag force is given by:

$D= C_D \frac{\rho}{2}V^2 A$

And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:

$D_1 = C_{D1} \frac{\rho}{2}V^2 (wh)$

Where w is the width and h the height.

The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:

$D_2 = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier})$

If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:

$\Delta P = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V- C_{D1} \frac{\rho}{2}V^2 (wh) V$

We can assume the same drag coeeficient $C_{D1}=C_{D2}=C_{D}$ and we got:

$\Delta P = C_{D} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V- C_{D} \frac{\rho}{2}V^2 (wh) V$

$\Delta P = C_{D} \frac{\rho}{2}V^3 (A_{carrier})$

1.7 ft =0.518 m

60 mph = 26.822 m/s

In order to find the drag coeffcient we ned to estimate the Reynolds number first like this:

$R_E= \frac{Vl}{v}= \frac{26.822m/s*0.518 m}{1.58x10^{-4} Pa s}= 8.79 x10^{4}$

And the value for the kinematic vicosity was obtained from the table of physical properties of the air under standard conditions.

Now we can find the aspect ratio like this:

$\frac{l}{h}=\frac{5}{1.7}2.941$

And we can estimate the calue of $C_D = 1.2$ from a figure.

And we can calculate the power difference like this:

$\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp$

8 0

2 years ago

Natural-draft furnaces are no longer available for new installations because they are _____.