4 years ago

Can anybody teach me how to make an app with flask and pygame together?

Engineering

1 answer:

sweet [91]4 years ago

8 0

Answer:

no :-(

Explanation:

Are you sure these technologies fit together?

As I understand it, flask is a library to write web services using python. It will essentially serve HTML pages and perhaps REST interfaces?

Pygame is a local machine abstraction layer to graphics and other i/o. I don't think it will run in a browser.

What is your plan?

You might be interested in

After being purged with nitrogen, a low-pressure tank used to store flammable liquids is at a total pressure of 0.03 psig. (a) I

g100num [7]

Answer:

See attached picture.

Explanation:

8 0

3 years ago

A circuit contains four 100 ohm resistors connected in series. If you test the circuit with a DMM you should read a total resist

olga nikolaevna [1]

C is the correct answer

8 0

3 years ago

What is the measurment unit of permeability?

nikitadnepr [17]

Answer:

Henries :)

Explanation:

I looked it up

7 0

3 years ago

Read 2 more answers

A W18 X 119 is used as a compression member with one end fixed and the other end pinned. The length is 12 feet. What is the desi

laiz [17]

Answer:

156.0 ksi

Explanation:

The formula of compressive strength is CS = F / A

where F is the force or load while A is the cross-sectional area

Given the inertia Ag = 35.1 in^4 , which is less than 40, the steel has a short column.

E = 29000 ksi

L = 12ft

r = 2.69 in

therefore;

CS = π^2 E / (kL^2/r)^2

= π^2 29000 / (0.8 x 12^2 / 2.69)^2

= 286292.756 / 1834.4089 = 156 ksi

3 0

3 years ago

How many joules are required to raise the temperature of a cubic meter of water by 10K?

kykrilka [37]

Answer:

4.186 × 10⁷ J

Explanation:

Heat gain by water = Q

Thus,

$m_{water}\times C_{water}\times \Delta T=Q$

For water:

Volume = 1 m³ = 1000 L ( as 1 m³ = 1000 L)

Density of water= 1 kg/L

So, mass of the water:

$Mass\ of\ water=Density \times {Volume\ of\ water}$

$Mass\ of\ water=1 kg/L \times {1000\ L}$

Mass of water = 1000 kg

Specific heat of water = 4.186 kJ/kg K

ΔT = 10 K

So,

$1000\times 4.186\times 10=Q$

Q = 41860 kJ

Also, 1 kJ = 1000 J

So, Q = 4.186 × 10⁷ J

8 0

4 years ago

Can anybody teach me how to make an app with flask and pygame together?​

Can anybody teach me how to make an app with flask and pygame together?