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aleksklad [387]
2 years ago
12

What are the disadvantages of military shovels?

Engineering
1 answer:
Xelga [282]2 years ago
4 0

Answer:

being prepared and having sufficient military resources available can be great IF there is an attack although it can be viewed as intimidating and scare off allies. also, it can cause other resources too thin.

Explanation:

it is great to have a sense of national pride and togetherness although if it gets out of hand it can be almost brainwashing and used to control citizens.

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Thin film deposition is a process where: a)-elemental, alloy, or compound thin films are deposited onto a bulk substrate! b)-Pho
marshall27 [118]

Answer:

(A) elemental, alloy, or compound thin films are deposited on to a bulk substrate

Explanation:

In film deposition there is process of depositing of material in form of thin films whose size varies between the nano meters to micrometers onto a surface. The material can be a single element a alloy or a compound.

This technology is very useful in semiconductor industries, in solar panels in CD drives etc

so from above discussion it is clear that option (a) will be the correct answer

8 0
3 years ago
A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is
-BARSIC- [3]

Answer:

\mathbf{C_{10} = 137.611 \ kN}

Explanation:

From the information given:

Life requirement = 40 kh = 40 40 \times 10^{3} \ h

Speed (N) = 520 rev/min

Reliability goal (R_D) = 0.9

Radial load (F_D) = 2600 lbf

To find C10 value by using the formula:

C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}

where;

x_D = \text{bearing life in million revolution} \\  \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}

\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}

The Weibull parameters include:

x_o = 0.02

(\theta - x_o) = 4.439

b= 1.483

∴

Using the above formula:

C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}

C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}

C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}

C_{10} = 30962.449 \ lbf

Recall that:

1 kN = 225 lbf

∴

C_{10} = \dfrac{30962.449}{225}

\mathbf{C_{10} = 137.611 \ kN}

7 0
2 years ago
Two steel plates are to be held together by means of 16-mm-diameter high-strength steel bolts fitting snugly inside cylindrical
dusya [7]

Answer:

The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm

Explanation:

For steel bolt

Stress = 210 MPa or 210 N/mm2

Pressure = Stress* Area

Pbolt = 210 N/mm2 * 16^2 *(pi)/4

Pbolt = 210 N/mm2 * 200.96 mm^2 = 42201.6  N

For Brass spacer

Pressure = 42201.6  N

Area of Brass spacer = Pressure/Stress

Area of Brass spacer = 42201.6  N/145 N/mm^2 = 291.044 mm^2

Area of Brass spacer = (pi) (d^2 - 16^2)/4 =  291.044 mm^2

d^2 - 16^2 = 291.044 mm^2* 4/(pi) = 370.758

d^2 =  370.758 + 16^2

d^2 =   626.758

d = 25.03 mm

The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm

5 0
3 years ago
Interpret the assembly program below: MOV R3,R0;
Reika [66]

Answer:

Explanation:

1.  With the operands R0, R1, the program would compute AND operation and ADD operation .

2. The operands could truly be signed 2's complement encoded (i.e Yes) .

3. The overflow truly occurs when two numbers that are unsigned were added and the result is larger than the capacity of the register, in that situation, overflow would occur and it could corrupt the data.

 When the result of an operation is smaller in magnitude than the smallest value represented by the data type, then arithmetic underflow will occur.

7 0
3 years ago
Ô tô có khối lượng m (kg) đặt tại trung tâm h . Khoảng cách từ h tới 2 bánh xe hai bên của a (m) và b (m) , khoảng cách vết bánh
Nutka1998 [239]

Answer:

wiwhwnwhwwbbwbwiwuwhwhehehewhehehheheheehehehehhehehwh

Explanation:

jwhwhwhwhwhwwhhahwhahahwh

6 0
3 years ago
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