Ask question

Ask question

All categories

3 years ago

7

Think about all the things we have investigated in class so far this semester. Identify three examples of climate-related aspect

s of the Earth system that we have discussed in the class.

1 answer:

Free_Kalibri [48]3 years ago

6 0

Answer:

1) the mean surface temperature of earth.

2) precipitation and sunshine

3) air and wind movement.

You might be interested in

Consider two different types of motors. Motor A has a characteristic life of 4100 hours (based on a MTTF of 4650 hours) and a sh

Daniel [21]

Answer:B

Explanation:

Given

For motor A

Characteristic life(r)=4100 hr

MTTF=4650 hrs

shape factor(B )=0.8

For motor B

Characteristic life(r)=336 hr

MTTF=300 hr

Shape Factor (B)=3

Reliability for 100 hours

$R_a=e^{-\left ( \frac{T-r}{n}\right )B}$

$R_a=e^{-\left ( \frac{4650-4100}{100}\right )0.8}$

$R_a=e^{-4.4}=0.01227$

For B

$R_b=e^{-\left ( \frac{300-336}{100}\right )3}$

$R_b=e^{1.08}=2.944$

B is better for 100 hours

(b)For 750 hours

$R_a=e^{-0.5866}=0.55621$

$R_b=e^{0.144}=1.154$

So here B is more Reliable.

3 0

3 years ago

A room has a width of 14.1 feet, a length of 15.5 feet, and a ceiling height of 12.0 ft. The average flow rate for this room's a

prohojiy [21]

Answer:

Your question lacks the time required hence i will calculate the Average flow rate using a general concept and an assumed time value of 25 seconds

ANSWER : 104.904 ft^3/sec

Explanation:

General concept : Average flow rate is the volume of fluid per unit time through an area

Hence the average flow rate of the air conditioning unit of this room

Volume of the room / time taken for the air to cycle the room = v / t

assuming the time taken = 25 seconds

volume of room = width * length * height

= 14.1 * 15.5 * 12 = 2622.6 ft^3

Average flow rate = V/ t

= 2622.6 / 25 = 104.904 ft^3/sec

8 0

3 years ago

I am trying to create a line of code to calculate distance between two points. (distance=[tex]\sqrt{ (x2-x1)^2+(y2-y1)^2}) My li

k0ka [10]

Answer:

point_dist = math.sqrt((math.pow(x2 - x1, 2) + math.pow(y2 - y1, 2))

Explanation:

The distance formula is the difference of the x coordinates squared, plus the difference of the y coordinates squared, all square rooted. For the general case, it appears you simply need to change how you have written the code.

point_dist = math.sqrt((math.pow(x2 - x1, 2) + math.pow(y2 - y1, 2))

Note, by moving the 2 inside of the pow function, you have provided the second argument that it is requesting.

You were close with your initial attempt, you just had a parenthesis after x1 and y1 when you should not have.

Cheers.

6 0

3 years ago

In your opinion...

ch4aika [34]

Answer:no

TTHANLS FOR FREE POINTS

Explanation:

8 0

2 years ago

Water at a pressure of 3 bars enters a short horizontal convergent channel at 3.5 m/s. The upstream and downstream diameters of

earnstyle [38]

Answer:

The pressure reduces to 2.588 bars.

Explanation:

According to Bernoulli's theorem for ideal flow we have

$\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=constant$

Since the losses are neglected thus applying this theorm between upper and lower porion we have

$\frac{P_{u}}{\gamma _{w}}+\frac{V-{u}^{2}}{2g}+z_{u}=\frac{P_{L}}{\gamma _{w}}+\frac{V{L}^{2}}{2g}+z_{L}$

Now by continuity equation we have

$A_{u}v_{u}=A_{L}v_{L}\\\\\therefore v_{L}=\frac{A_{u}}{A_{L}}\times v_{u}\\\\v_{L}=\frac{d^{2}_{u}}{d^{2}_{L}}\times v_{u}\\\\\therefore v_{L}=\frac{2500}{900}\times 3.5\\\\\therefore v_{L}=9.72m/s$

Applying the values in the Bernoulli's equation we get

$\frac{P_{L}}{\gamma _{w}}=\frac{300000}{\gamma _{w}}+\frac{3.5^{2}}{2g}-\frac{9.72^{2}}{2g}(\because z_{L}=z_{u})\\\\\frac{P_{L}}{\gamma _{w}}=26.38m\\\\\therefore P_{L}=258885.8Pa\\\\\therefore P_{L}=2.588bars$

6 0

3 years ago