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2 years ago

How many 3-digits numbers which are greater than 300 can be formed from 6

Engineering

1 answer:

Marizza181 [45]2 years ago

6 0

Explanation:

111 even numbers greater than 300 can be formed with the digits 1,2,3,4,5 if repetition of digits in a number is not allowed.

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Using an "AND" and an "OR", list all information (Equipment Number, Equipment Type, Seat Capacity, Fuel Capacity, and Miles per

Tomtit [17]

Answer:

Explanation :

The given information to be listed can are Equipment Number, Equipment Type, Seat Capacity, Fuel Capacity, and Miles per Gallon.

Check the attached document for the solution.

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2 years ago

Shows a closed tank holding air and oil to which is connected a U-tube mercury manometer and a pressure gage. Determine the read

damaskus [11]

Answer:

$P_2-P_1=27209h$

Explanation:

For pressure gage we can determine this by saying:

The closed tank with oil and air has a pressure of P₁ and the pressure of oil at a certain height in the U-tube on mercury is p₁gh₁. The pressure of mercury on the air in pressure gauge is p₂gh₂. The pressure of the gage is P₂.

$P_1+p_1gh_1=p_2_gh_2+P_2$

We want to work out P₁-P₂: Heights aren't given so we can solve it in terms of height: assuming h₁=h₂=h

$P_1-P_2=p_1gh_1-p_2gh_2=(55)\cdot{32.2}h-845\cdot{32.2}h$

$P_2-P_1=27209h$

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3 years ago

How to study thermodynamics?

expeople1 [14]

It is study of the relationships between heat, temprature, work and energy

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2 years ago

Read 2 more answers

A gas tank is known to have a thickness of 0.5 inches and an internal pressure of 2.2 ksi. Assuming that the maximum allowable s

sergiy2304 [10]

Answer:

$D_o=11.9inch$

Explanation:

From the question we are told that:

Thickness $T=0.5$

Internal Pressure $P=2.2Ksi$

Shear stress $\sigma=12ksi$

Elastic modulus $\gamma= 35000$

Generally the equation for shear stress is mathematically given by

$\sigma=\frac{P*r_1}{2*t}$

Where

r_i=internal Radius

Therefore

$12=\frac{2.2*r_1}{2*0.5}$

$r_i=5.45$

Generally

$r_o=r_1+t$

$r_o=5.45+0.5$

$r_o=5.95$

Generally the equation for outer diameter is mathematically given by

$D_o=2r_o$

$D_o=11.9inch$

Therefore

Assuming that the thin cylinder is subjected to integral Pressure

Outer Diameter is

$D_o=11.9inch$

7 0

2 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:

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