All categories

2 years ago

3. What is a caliber (relate it to rockets)

Engineering

1 answer:

Kamila [148]2 years ago

6 0

Answer:

In this context a caliber is defined as the diameter of the body tube, and it is used to support the general rule of thumb that for a rocket of typical aspect ratio to be stable the CG should be one caliber ahead of CP.

Explanation:

You might be interested in

What have you learned about designing solutions? How does this apply to engineering? Think of some engineering solutions that st

Andrew [12]

Answer:

In engineering design, failure is expected. It helps you find the best solutions before implementing them in the “real world”. Having a prototype fail is a GOOD thing, because that means you have learned something new about the problem and potential solutions.

Explanation:

4 0

2 years ago

For the pipe-fl ow-reducing section of Fig. P3.54, D 1 5 8 cm, D 2 5 5 cm, and p 2 5 1 atm. All fl uids are at 20 8 C. If V 1 5

bonufazy [111]

Answer:

The total force resisted by the flange bolts is 163.98 N

Explanation:

Solution

The first step is to find the pipe cross section at the inlet section

Now,

A₁ = π /4 D₁²

D₁ = diameter of the pipe at the inlet section

Now we insert 8 cm for D₁ which gives us A₁ = π /4 D (8)²

=50.265 cm² * ( 1 m²/100² cm²)

= 5.0265 * 10^⁻³ m²

Secondly, we find cross section area of the pipe at the inlet section

A₂ = π /4 D₂²

D₂ = diameter of the pipe at the inlet section

Now we insert 5 cm for D₁ which gives us A₁ = π /4 D (5)²

= 19.63 cm² * ( 1 m²/100² cm²)

= 1.963 * 10^⁻³ m²

Now,

we write down the conversation mass relation which is stated as follows:

Q₁ = Q₂

Where Q₁ and Q₂ are both the flow rate at the exist and inlet.

We now insert A₁V₁ for Q₁ and A₂V₂ for Q₂

So,

V₁ and V₂ are defined as the velocities at the inlet and exit

We now insert 5.0265 * 10^⁻³ m² for A₁ 5 m/s for V₁ and 1.963 * 10^⁻³ m² for A₂

= 5.0265 * 5 = 1.963 * V₂

V₂ = 12.8 m/s

Note: Kindly find an attached copy of the part of the solution to the given question below

8 0

2 years ago

In a creep test, increasing the temperature will (choose the best answer) A. increase the instantaneous initial deformation B. i

Hitman42 [59]

Answer:

All of the above

Explanation:

firstly, a creep can be explained as the gradual deformation of a material over a time period. This occurs at a fixed load with the temperature the same or more than the recrystallization temperature.

Once the material gets loaded, the instantaneous creep would start off and it is close to electric strain. in the primary creep area, the rate of the strain falls as the material hardens. in the secondary area, a balance between the hardening and recrystallization occurs. The material would get to be fractured hen recrstallization happens. As temperature is raised the recrystallization gets to be more.

8 0

2 years ago

A rectangular open box, 25 ft by 10 ft in plan and 12 ft deep weighs 40 tons. Sufficient amount of stones is placed in the box a

dimulka [17.4K]

Answer:

44.95 tonnes

Explanation:

According to principle of buoyancy the object will just sink when it's weight is more than the weight of the liquid it displaces

It is given that empty weight of box = 40 tons

Let the mass of the stones to be placed be = M tonnes

Thus the combined mass of box and stones = (40+M) tonnes..........(i)

Since the box will displace water equal to it's volume V we have $volume of box = 25ft*10ft*12ft= 3000ft^{3}$