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2 years ago

What is mechanical engineer

Engineering

1 answer:

Charra [1.4K]2 years ago

4 0

A man or women who design other machines inside building, etc. pls give brainliest

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Voltage drop testing is being discussed.

Airida [17]

Answer:

Technician B only is correct

Explanation:

Voltage drop testing is a method used to find the amount of electrical resistance available in an high amperage circuit that involves connecting the leads of the meter in parallel to the circuits being tested such that disassembly is not required

In voltage drop test, the red voltmeter lead and black voltmeter lead are placed at two points on the same side of the circuit connection such that the leads are in between two positive connection or two negative connection (rather than connecting the red to the positive and the black to the negative sides of the circuit) and digital voltmeter is used for the voltage drop measurement across the lead while the connection is under load.

Therefore, Technician B only is correct

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2 years ago

State two faults that are common in a simple cell

Step2247 [10]

Answer:

the two defects of a simple cell are:

1. Polarization

2. Local action

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2 years ago

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Please help!!!! WHICH 2 SKILLS N ABILITIES R ESSENTIAL FOR A SHIP CAPTAIN?? A. Good vision B. Public speaking skills C. Leadersh

Brilliant_brown [7]

A and C are the correct answers

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3 years ago

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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:

$\sum M_A = 0$