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damaskus [11]
3 years ago
15

Give me source code of Simple openGL project. ( without 3D or Animation) simple just.

Engineering
1 answer:
Ivan3 years ago
4 0

Answer:

Use GitHub or stackoverflow for this answer

Explanation:

It helps with programming a lot

You might be interested in
1: asha started abusness with 30.000
svetoff [14.1K]

Answer:

Explanation:adrive with visual acutity of 20/30 can just decipher asing adistance 20ft from asing determine the maximum destance from the sing which drivers with the flowing visual acuities will able to see the same sing 20/15 20/50

4 0
2 years ago
Water is pumped steadily through a 0.10-m diameter pipe from one closed pressurized tank to another tank. The pump adds 4.0 kW o
jekas [21]

Complete Question

Complete Question is attached below.

Answer:

V'=5m/s

Explanation:

From the question we are told that:

Diameter d=0.10m

Power P=4.0kW

Head loss \mu=10m

 \frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu

 \frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10

 H_m=(\frac{200*10^3}{1000*9.8}-10)

 H_m=10.39m

Generally the equation for Power is mathematically given by

 P=\rho gQH_m

Therefore

 Q=\frac{P}{\rho g H_m}

 Q=\frac{4*10^4}{1000*9.81*10.9}

 Q=0.03935m^3/sec

Since

 Q=AV'

Where

 A=\pi r^2\\A=3.142 (0.05)^2

 A=7.85*10^{-3}

Therefore

 V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}

 V'=5m/s

5 0
3 years ago
What’s the number of gold atoms in a nanogram? a picogram?
zvonat [6]

Answer :

The number of gold atoms in nanogram is, 3.057\times 10^{12}

The number of gold atoms in picogram is, 3.057\times 10^{9}

Explanation :

As we know that the molar mass of gold is, 196.97 g/mole. That means, 1 mole of gold has 196.97 grams of mass of gold.

As we know that,

1 mole contains 6.022\times 10^{23} number of atoms.

First we have to determine the number of gold atoms in a nanogram.

As, 196.97 grams of gold contains 6.022\times 10^{23} number of gold atoms

And, 1 grams of gold contains \frac{1g}{196.97g}\times (6.022\times 10^{23}) number of gold atoms

So, 10^{-9} nanograms of gold contains \frac{1g}{196.97g}\times (10^{-9})\times (6.022\times 10^{23})=3.057\times 10^{12} number of gold atoms

The number of gold atoms in nanogram is, 3.057\times 10^{12}

Now we have to determine the number of gold atoms in a picogram.

As, 196.97 grams of gold contains 6.022\times 10^{23} number of gold atoms

And, 1 grams of gold contains \frac{1g}{196.97g}\times (6.022\times 10^{23}) number of gold atoms

So, 10^{-12} picograms of gold contains \frac{1g}{196.97g}\times (10^{-12})\times (6.022\times 10^{23})=3.057\times 10^{9} number of gold atoms

The number of gold atoms in picogram is, 3.057\times 10^{9}

8 0
3 years ago
Liquid ethanol is a flammable fluid and can release vapors that form explosive mixtures at temperatures above its flashpoint at
marta [7]

Answer:

The volume flow rate necessary to keep the temperature of the ethanol in the pipe below its flashpoint should be greater than 1.574m^3/s

Explanation:

Q = MCp(T2 - T1)

Q (quantity of heat) = Power (P) × time (t)

Density (D) = Mass (M)/Volume (V)

M = DV

Therefore, Pt = DVCp(T2 - T1)

V/t (volume flow rate) = P/DCp(T2 - T1)

P = 20kW = 20×1000W = 20,000W, D(rho) = 789kg/m^3, Cp = 2.44J/kgK, T2 = 16.6°C = 16.6+273K = 289.6K, T1 = 10°C = 10+273K = 283K

Volume flow rate = 20,000/789×2.44(289.6-283) = 20,000/789×2.44×6.6 = 1.574m^3/s (this is the volume flow rate at the flashpoint temperature)

The volume flow rate necessary to keep the ethanol below its flashpoint temperature should be greater than 1.574m^3/s

6 0
3 years ago
5. The water in an 8-m-diameter, 3-m-high above-ground swimming pool is to be emptied by unplugging a 3-cm-diameter, 25-m-long h
frosja888 [35]

Answer:

The maximum discharge rate of water through the pipe is 0.00545 m³/s or 5.45 L/s.

Friction head and pressure head will cause the actual flow rate to be less.

Explanation:

Considering point 1 at the free surface of the pool, and point 2 at the exit of

pipe.

Using Bernoulli equation between

these two points simplifies to

P1/(p*g) + V1²/2g + z1 = P2/(p*g) + V2²/2g + z2

Let the reference level at the pipe exit (z2 = 0). Noting that the fluid at both points is open to the atmosphere (and thus P1 = P2 = Patm) and that the fluid velocity at the free surface is very low (V1 ≅ 0),

P/(p*g) + z1 = P/(p*g) + V2²/2g

z1 = V2²/2g

Note; z1 = h

V2max = √2gh

h = 3 m

V2max = √2 * 9.81 * 3

V2max = √58.86 = 7.67 m/s

maximum discharge rate of water through the pipe Qmax = Area A * Velocity of discharge V2max

Qmax = A * V2max

Diameter d = 3 cm = 0.03 m

A = Πd²/4 = (Π * 0.03²)/4 = 0.00071m³

Qmax = 0.00071 * 7.67 = 0.00545 m³/s

Qmax = 5.45 L/s

The maximum discharge rate of water through the pipe is 0.00545 m³/s or 5.45 L/s.

Actual flow rate will be less because of heads such as friction head and pressure head.

7 0
3 years ago
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