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oksano4ka [1.4K]
2 years ago
5

REVVIVE ME MY MOM WALKED IN MY ROOM AND SCARED THE BAJESUS OUTTA ME

Engineering
2 answers:
bagirrra123 [75]2 years ago
7 0
Ok *revives* r u ok now
vlada-n [284]2 years ago
5 0

Answer: The answer to this question is tell your mother I thought you were someone else...

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What is a motor cycle motor made out of
Natali5045456 [20]

Explanation:

a motorcycle Motor is made out of iron

4 0
3 years ago
Read 2 more answers
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
All of these are uses of microwaves except...
Talja [164]

Answer:D

Explanation:

5 0
3 years ago
Consider a Carnot heat pump cycle executed in a steady-flow system in the saturated mixture region using R-134a flowing at a rat
attashe74 [19]

Answer:

7.15

Explanation:

Firstly, the COP of such heat pump must be measured that is,

              COP_{HP}=\frac{T_H}{T_H-T_L}

Therefore, the temperature relationship, T_H=1.15\;T_L

Then, we should apply the values in the COP.

                           =\frac{1.15\;T_L}{1.15-1}

                           =7.67

The number of heat rejected by the heat pump must then be calculated.

                   Q_H=COP_{HP}\times W_{nst}

                          =7.67\times5=38.35

We must then calculate the refrigerant mass flow rate.

                   m=0.264\;kg/s

                   q_H=\frac{Q_H}{m}

                         =\frac{38.35}{0.264}=145.27

The h_g value is 145.27 and therefore the hot reservoir temperature is 64° C.

The pressure at 64 ° C is thus 1849.36 kPa by interpolation.

And, the lowest reservoir temperature must be calculated.

                   T_L=\frac{T_H}{1.15}

                        =\frac{64+273}{1.15}=293.04

                        =19.89\°C

the lowest reservoir temperature = 258.703  kpa                    

So, the pressure ratio should be = 7.15

8 0
3 years ago
The local atmospheric pressure is measured with a water barometer. If the water column is measured to be 30 ft, what is the atmo
ella [17]

Answer:

The atmospheric pressure in atm=0.885 atm

Explanation:

Given that

Local pressure (h)= 30 ft of water height     ( 1 ft= 0.3048 m)

We know that pressure in given by

  P=ρgh

We know that ρ of water is 1000\dfrac{kg}{m^3}

So pressure

P=1000(9.81)(9.144)

P=89.7026 \dfrac{N}{m^2}

We know that 1000 Pa=0.00986 atm

So P=0.885 atm

The atmospheric pressure in atm=0.885 atm

8 0
3 years ago
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