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Oliga [24]
3 years ago
11

A vehicle has an oil leak that is causing the entire oil pan to be wet, but inspection reveals no exact source after cleaning of

f the oil residue. technician a says to install a fluorescent dye in the crankcase and operate the engine, then re-inspect for leaks with a special light (black light). technician b says the oil leak may be coming from a source at the top of the engine, such as a valve cover gasket. who is correct?
Physics
1 answer:
attashe74 [19]3 years ago
7 0

Answer:

Technician A is correct

Explanation:

The best approach to solve the problem is that of technician A. using a fluorescent die is the easiest and most efficient way to trace leaks with unknown sources. The fluorescent die will simply illuminate the path to the leaking spot in the engine of the car, without any need for much speculations. This makes this method a sure approach.

However, Technician B's approach still has a lot of assumptions factored into the methodology, and would not work properly. It will still require the painstaking attempts trying to make guesses where the oil leak is coming from, which will lead to wastage of time and energy.

This makes Technician A have the right approach to solving the problem

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a 1500 kg car accelerates uniformly from rest to 10.0 meters per secound in 3.0 secound .what is the work done on the car in thi
zubka84 [21]

Answer:

The work done on the car is, W = 75,000 J

The power delivered by the engine, P = 25,000 watts

Explanation:

Given,

The mass of the car, m = 1500 Kg

The initial velocity of the car, u = 0

The final velocity of the car, v = 10 m

The time duration of the travel, t = 3 s

Using the first equation of motion

                     v = u + at

                     a = (v - u) / t

Substituting the given values in the above equation

                    a = (10 - 0) / 3

                       = 3.33 m/s²

Using the second equations of motion

                      s = ut + 1/2 at²

                         = 0 + 0.5 x 3.33 x 3²

                         = 15 m

The force exerted by the car

                         F = m x a

                            = 1500 Kg x 3.33 m/s²

                            = 5000 N

The work done by the car,

                          W = F x S

                               = 5000 N x 15 m

                               = 75,000 J

Hence, the work done on the car is, W = 75,000 J

The power delivered by the engine,

                            P = W / t

                                = 75,000 J / 3 s

                                 = 25,000 watts

The power delivered by the engine, P = 25,000 watts

5 0
4 years ago
The SI system uses three base units. Is this true or false?
ozzi
The\text{ SI }system\text{ uses seven }base\text{ units, hence the statement is false}

8 0
1 year ago
Read 2 more answers
Can an element be a molecule
Serga [27]

an element can make a molecule. so technically yes.

6 0
3 years ago
Read 2 more answers
A traffic light of weight 100 N is supported by two ropes that make 30-degree angle with the horizontal. What is the vertical co
alina1380 [7]

Answer:

T_{vertical} = 50 N

Explanation:

given,

traffic light weight = 100 N

angle at which the rope is supported = 30°

vertical component of force = ?

2 T sin \theta= W

2 T sin 30^0= 100

T = \dfrac{100}{2 sin 30^0}

T = 100\ N

T_{vertical} = T sin 30^0

T_{vertical} = 100\times \dfrac{1}{2}

T_{vertical} = 50 N

5 0
4 years ago
8. A rectangle is measured to be 6.4 +0.2 cm by 8.3 $0.2 cm.
mamaluj [8]

Answer:

a) The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty in its perimeter is 0.8 centimeters.

Explanation:

a) From Geometry we remember that the perimeter of the rectangle (p), measured in centimeters, is represented by the following formula:

p = 2\cdot (w+l) (1)

Where:

w - Width, measured in centimeters.

l - Length, measured in centimeters.

If we know that w = 6.4\,cm and l = 8.3\,cm, then the perimeter of the rectangle is:

p = 2\cdot (6.4\,cm+8.3\,cm)

p = 29.4\,cm

The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty of the perimeter (\Delta p), measured in centimeters, is estimated by differences. That is:

\Delta p = 2\cdot (\Delta w + \Delta l)  (2)

Where:

\Delta w - Uncertainty in width, measured in centimeters.

\Delta l - Uncertainty in length, measured in centimeters.

If we know that \Delta w = 0.2\,cm and \Delta l = 0.2\,cm, then the uncertainty in perimeter is:

\Delta p = 2\cdot (0.2\,cm+0.2\,cm)

\Delta p = 0.8\,cm

The uncertainty in its perimeter is 0.8 centimeters.

5 0
3 years ago
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