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kvasek [131]
3 years ago
13

The current standard for recovery only equipment is sae

Engineering
1 answer:
vodomira [7]3 years ago
7 0

Explanation:

I think is SAE standard J2810

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Please help me with this, picture.
Alenkasestr [34]
Maybe try 086 degrees
3 0
3 years ago
Transcript
posledela

Answer:

O is truse is the best answer hhahahha

Explanation:

8 0
3 years ago
Each cout statement has a syntax error. Type the first cout statement, and press Run to observe the error message. Fix the error
saul85 [17]

Answer:

1. cout << "Num: " << songNum << endl;

2. cout << songNum << endl;

3. cout << songNum <<" songs" << endl;

Explanation:

//Full Code

#include <iostream>

using namespace std;

int main ()

{

int songNum;

songNum = 5;

cout << "Num: " << songNum << endl;

cout << songNum << endl;

cout << songNum <<" songs" << endl;

return 0;

}

1. The error in the first cout statement is that variable songnum is not declared.

C++ is a case sensitive programme language; it treats upper case and lower case characters differently.

Variable songNum was declared; not songnum.

2. Cout us used to print a Variable that has already been declared.

The error arises in int songNum in the second cout statement.

3. When printing more than one variables or values, they must be separated with <<

4 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
3 years ago
Write a class named FBoard for playing a game, where player x is trying to get her piece to row 7 and player o is trying to make
OverLord2011 [107]

Answer:

12345=V

Explanation:

7 0
4 years ago
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