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r-ruslan [8.4K]
3 years ago
13

C programming fundamentals for everyone​

Engineering
1 answer:
densk [106]3 years ago
6 0

Answer:I am so sorry but here are no questions, thanks for the points

Explanation:Have a nice day

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Software that is released to have users test out the "bugs" is known as Ransomeware O Break-in software 2 O Flim flam software O
Sophie [7]

Answer:

Beta software

Explanation:

5 0
3 years ago
Calculate the number of atoms per cubic meter in Metal B (units atoms/m^3). Write your answer with 4 significant figures metal:
Iteru [2.4K]

Answer:

7,217*10^28 atoms/m^3

Explanation:

  • Metal: Vanadium
  • Density: 6.1 g/cm^3
  • Molecuar weight: 50,9 g/mol

The Avogadro's Number, 6,022*10^23, is the number of atoms in one mole of any substance. To calculate the number of atoms in one cubic meter of vanadium we write:

1m^3*(100^3 cm^3/1 m^3)*(6,1 g/1 cm^3)*(1 mol/50,9g)*(6,022*10^23 atoms/1 mol)=7,217*10^28 atoms

Therefore, for vanadium we have 7,217*10^28 atoms/m^3

6 0
3 years ago
A 132mm diameter solid circular section​
Ganezh [65]

Answer:

not sure if this helps but

5 0
3 years ago
A medium-sized jet has a 3.8-mm-diameter fuselage and a loaded mass of 85,000 kg. The drag on an airplane is primarily due to th
SCORPION-xisa [38]

Answer:

F_{thrust} ≅ 111 KN

Explanation:

Given that;

A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8

mass = 85,000 kg

drag co-efficient (C) = 0.37

(velocity (v)= 230 m/s

density (ρ) = 1.0 kg/m³

To calculate the thrust; we need to determine the relation of the drag force; which is given as:

F_{drag} = \frac{1}{2} × CρAv²

where;

ρ = density of air wind.

C = drag co-efficient

A = Area of the jet

v = velocity of the jet

From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0

SO, F_{drag}-F_{thrust} = 0

We can as well say:

F_{drag}= F_{thrust}

We can now replace F_{thrust} with F_{drag} in the above equation.

Therefore, F_{thrust} = \frac{1}{2} × CρAv²

The A which stands as the area of the jet is given by the formula:

A=\frac{\pi d^2}{4}

We can now have a new equation after substituting our A into the previous equation as:

F_{thrust} = \frac{1}{2} × Cρ (\frac{\pi d^2}{4})v^2

Substituting our data from above; we have:

F_{thrust} = \frac{1}{2} × (0.37)(1.0kg/m^3)(\frac{\pi(3.8m)^2 }{4})(230m/s)^2

F_{thrust} = \frac{1}{8}   (0.37)(1.0kg/m^3)({\pi(3.8m)^2 })(230m/s)^2

F_{thrust} = 110,990N

F_{thrust}  in N (newton) to KN (kilo-newton) will be:

F_{thrust} = (110,990N)*\frac{1KN}{1,000N}

F_{thrust} = 110.990 KN

F_{thrust} ≅ 111 KN

In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.

5 0
3 years ago
A water supply agency is planning to add two reservoirs to its system. Water will flow from Reservoir A to Reservoir B via a 10,
NikAS [45]

Attached is the solution to the above question.

3 0
3 years ago
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