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lora16 [44]
3 years ago
6

View the picture below and then correctly answer the questions using the following words: Temperate Zone, Tropical Zone, Polar Z

one, Polar Winds, Westerly Winds, Trade Winds.
Use the following words to answer the 5 questions below. Temperate Zone, Tropical Zone, Polar Zone, Polar Winds, Westerly Winds, Trade Winds.

1. What winds blow in the zone labeled "B"?

2. What winds blow in the zone labeled "C"?

3. What is the name of the zone labeled "A"?

4. The winds blow in the zone labeled "A"?

5. What is the name of the zone labeled "B"?
Engineering
2 answers:
Nadusha1986 [10]3 years ago
8 0

Can you provide the picture? Thanks !

Otrada [13]3 years ago
7 0
C ITS DEFINITELY C C C C
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Write a Nested While Loop that will increment the '*' from 1 to 10.
andrey2020 [161]

Answer:

The program is as follows:

i = 1

while(i<11):

   j = 1

   while(j<=i):

       print('*', end = '')

       j += 1

   i += 1

   print()

Explanation:

Initialize i to 1

i = 1

The outer loop is repeated as long as i is less than 11

while(i<11):

Initialize j to 1

   j = 1

The inner loop is repeated as long as j is less than or equal i

   while(j<=i):

This prints a *

       print('*', end = '')

This increments j and ends the inner loop

       j += 1

This increments i

   i += 1

This prints a blank and ends the inner loop

   print()

8 0
3 years ago
A hypothetical A-B alloy of composition 57 wt% B-43 wt% A at some temperature is found to consist of mass fractions of 0.5 for b
Dennis_Churaev [7]

Answer:

composition of alpha phase is 27% B

Explanation:

given data

mass fractions  = 0.5 for both

composition = 57 wt% B-43 wt% A

composition = 87 wt% B-13 wt% A

solution

as by total composition Co = 57 and by beta phase composition  Cβ = 87  

we use here lever rule that is

Wα = Wβ   ...............1

Wα = Wβ = 0.5

now we take here left side of equation

we will get

\frac{C_\beta - Co}{C_\beta - Ca}   = 0.5

\frac{87 - 57}{87 - Ca} = 0.5  

solve it we get

Ca = 27

so composition of alpha phase is 27% B

8 0
3 years ago
An electric motor is to be supported by four identical mounts. Each mount can be treated as a linear prevent problems due requir
Artyom0805 [142]

GIVEN:

Amplitude, A = 0.1mm

Force, F =1 N

mass of motor, m = 120 kg

operating speed, N = 720 rpm

\frac{A}{F} =  \frac{0.1\times 10^{-3}}{1} = 0.1\times 10^{-3}

Formula Used:

A = \frac{F}{\sqrt{(K_{t} - m\omega ^{2}) +(\zeta \omega ^{2})}}

Solution:

Let Stiffness be denoted by 'K' for each mounting, then for 4 mountings it is 4K

We know that:

\omega = \frac{2 \pi\times N}{60}

so,

\omega = \frac{2 \pi\times 720}{60} = 75.39 rad/s

Using the given formula:

Damping is negligible, so, \zeta = 0

\frac{A}{F} will give the tranfer function

Therefore,

\frac{A}{F} = \frac{1}{\sqrt{(4K - 120\ ^{2})}}

0.1\times 10^{-3} =  \frac{1}{\sqrt{(4K - 120\ ^{2})}}

Required stiffness coefficient, K = 173009 N/m = 173.01 N/mm

8 0
3 years ago
A gas turbine operates with a regenerator and two stages of reheating and intercooling. Air enters this engine at 14 psia and 60
Rzqust [24]

Answer:

flow(m) = 7.941 lbm/s

Q_in = 90.5184 Btu/lbm

Q_out = 56.01856 Btu/lbm

Explanation:

Given:

- T_1 = 60 F = 520 R

- T_6 = 940 = 1400 R

- Heat ratio for air k = 1.4

- Compression ratio r = 3

- W_net,out = 1000 hp

Find:

mass flow rate of the air

rates of heat addition and rejection

Solution:

- Using ideal gas relation compute T_2, T_4, T_10:

                     T_2 = T_1 * r^(k-1/k)

                     T_2 = T_4 = T_10 = 520*3^(.4/1.4) = 711.744 R

- Using ideal gas relation compute T_7, T_5, T_9:

                     T_7 = T_6 * r^(-k-1/k)

                     T_7 = T_5 = T_9 = 1400*3^(-.4/1.4) = 1022.84 R

- The mass flow rate is obtained by:

                     flow(m) = W_net,out / 2*c_p*(1400-1022.84-711.744+520)

                     flow(m) = 1000*.7068 / 2*0.24*(1400-1022.84-711.744+520)

                     flow(m) = 7.941 lbm/s

- The heat input is as follows:

                     Q_in = c_p*(T_6 - T_5)

                     Q_in = 0.24*(1400 - 1022.84)

                     Q_in = 90.5184 Btu/lbm

- The heat output is as follows:

                     Q_out = c_p*(T_10 - T_1)

                     Q_out = 0.24*(711.744 - 520)

                    Q_out = 56.01856 Btu/lbm

                                           

                     

5 0
3 years ago
How does a catapult incorporate technology?
storchak [24]
I’m sure the answer is correct and it is below
“ A catapult works because energy can be converted from one type to another and transferred from one object to another. ... This energy is stored in the launching device as potential, or stored, energy. The catapult you are about to make uses elastic potential energy stored in a wooden stick as you bend it.”
6 0
3 years ago
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