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3 years ago

Identify three operational controls and explain how to use them?

Engineering

1 answer:

ladessa [460]3 years ago

8 0

Answer:

5.1 Personnel Security. ...

5.2 Physical and Environmental Protection. ...

5.3 Production, Input and Output Controls. ...

5.4 Contingency Planning and Disaster Recovery. ...

5.5 System Configuration Management Controls. ...

5.6 Data Integrity / Validation Controls. ...

5.7 Documentation. ...

5.8 Security Awareness and Training.

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The primary heat transfer mechanism that quickly warms my hand if I hold it directly above a campfire is: a)-Radiation b)-Induct

Tanzania [10]

Answer:

The correct answer is option 'c':Convection.

Explanation:

When we ignite a campfire the heat produced by combustion heats the air above the fire. As we know that if a gases gains heat it expands thus it's density decreases and hence it rises, if we hold our hands directly above the fire this rising hot air comes in contact with our hands thus warming them.

The situation is different if we are at some distance from the campfire laterally. Since the rising air cannot move laterally the only means the heat of the fire reaches our body is radiation.

But in the given situation the correct answer is convection.

6 0

3 years ago

A tool chest has 650 N weight that acts through the midpoint of the chest. The chest is supported by feet at A and rollers at B.

erica [24]

Answer:

the value of horizontal force P is 170.625 N

the value of horizontal force at P = 227.5 N is that the block moves to right and this motion is due to sliding.

Explanation:

The first diagram attached below shows the free body diagram of the tool chest when it is sliding.

Let start out by calculating the friction force

$F_f= \mu N_2$

where :

$F_f =$ friction force

$\mu$ = coefficient of friction

$N_2$ = normal friction

Given that:

$\mu$ = 0.3

$F_f =$ 0.3 $N_2$

Using the equation of equilibrium along horizontal direction.

$\sum f_x = 0$

P - $F_f =$ 0

P = 0.3 $N_2$ ----- Equation (1)

To determine the moment about point B ; we have the expression

$\sum M_B = 0$

0 = $N_2*70-W*35-P*100$

where;

P = horizontal force

$N_2$ = normal force at support A

W = self- weight of tool chest

Replacing W = 650 N

0 = $N_2*70-650*35-100*P$

$P = \frac{70 N_2-22750}{100} ----- equation (2)$

Replacing $\frac{70 N_2-22750}{100}$ for P in equation (1)

$\frac{70N_2 -22750}{100} =0.3 N_2$

$N_2 = \frac{22750}{40}$ $N_2 = 568.75 \ N$

Plugging the value of $N_2 = 568.75 \ N$ in equation (2)

$P = \frac{70(568.75)-22750}{100} \\ \\ P = \frac{39812.5-22750}{100} \\ \\ P = \frac{17062.5}{100}$

P =170.625 N

Thus; the value of horizontal force P is 170.625 N

b) From the second diagram attached the free body diagram; the free body diagram of the tool chest when it is tipping about point A is also shown below:

Taking the moments about point A:

$\sum M_A = 0$