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Vlada [557]
3 years ago
15

Technician A says that low pressure smoke installed in the fuel system can be used to check for leaks. Technicians B says that n

itrogen under low pressure can be installed in the fuel system to check for leaks. Which technician is correct?A. Technician A only.B. Technician B only. C. Technician A and B. D. D. Neither Techncian A and B.
Physics
1 answer:
pickupchik [31]3 years ago
4 0

Answer:

A. Technician A only.B.

Explanation: The fuel system of a vehicle is made up of the fuel pump,the fuel filter,the injector or carburettor and the fuel tank. The main function of the fuel system is supply fuel to the engine of a vehicle. In order to check for leakage in the fuel system a small pressure smoke is usually installed in the fuel system.

Nitrogen at low pressure has also been used to check for leakage in the fuel system of a vehicle.

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Suppose you have measured the diffraction pattern of a grating with d = 0.19 mm and have found that the spots were separated by
cestrela7 [59]

Answer:

d = 0.076 mm

Explanation:

Given data

diffraction pattern d1 = 0.19 mm = 0.019 cm

separated s(1)  = 1.8 cm

separated s(2) = 4.5 cm

to find out

d2 for an unknown

solution

we know here that spacing in between the diffraction fringe is always inversely proportional to diffraction grating so

we will apply here formula for unknown d that is

d1 (s1 / L) = d2 (s2 /L)

d2  = d1 ×  s(1) / s(2)

put here all thes evalue we get d2

d2  = d1 ×  s(1) / s(2)

d2  = 0.019 ×  1.8  / 4.5

d2 = 0.0076 cm

d2 = 0.076 mm

7 0
3 years ago
if a string of light goes out when one of the bulbs is removed, are the lights probably connected in a series circuit or a paral
Minchanka [31]
Yes it is because other wise the light would stay on
4 0
3 years ago
Read 2 more answers
A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.820 m. W
Vikentia [17]

Answer:

1.0752 kgm/s

Explanation:

Considering when the drop was dropped from rest from a height,

mass of the ball, m = 0.120 kg

height, h = - 1.25 m

the initial velocity, u = 0 m/s

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

                            V^{2} = U^{2} + 2gh

Substituting the values,

                             V^{2} = 0^{2} + 2(-9.8 m/s^{2})(-1.25 m)

                             V^{2} = 24.5 m/s

                             V = \sqrt{24.5} \ m/s

                             V = 4.95 \ m/s

                            V = ± 4.95 m/s

                            V = - 4.95 m/s

Since the ball is moving downward, the final velocity of the ball when it hits the floor is  V = - 4.95 m/s  

Considering when the ball rebounds from the floor,

assume the mass of the ball still remain, m = 0.120 kg

height, h = 0.820 m

the final velocity, v = 0 m/s  

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

                            V^{2} = U^{2} + 2gh

Substituting the values,

                            0^{2} = U^{2} + 2(-9.8 m/s^{2})(0.820 m)

                            0 = U^{2} - 16.072 m/s

                            U^{2} = 16.072 m/s

                            U = \sqrt{16.072} \ m/s

                           U = ± 4.01 m/s

                          U = + 4.01 m/s

Since the ball is moving upward, the initial velocity of the ball from the bounce from the floor is  U = + 4.01 m/s                        

From Newton's second law of motion, applied force is directly proportional to the rate of change in momentum.

                            F = \frac{mv - mu}{t}

                          F.t = m(v - u)

       ⇒      Impulse = Change in momentum

To calculate the impulse, the moment before the ball hits the ground will be the initial momentum while the moment the ball rebounces will be the final velocity,                        

          ∴          F.t = 0.120  kg(4.01  m/s - (-4.95  m/s) )

                      F.t = 0.120  kg(4.01  m/s + 4.95  m/s) )

                      F.t = 0.120  kg × 8.96  m/s

                      Impulse  = 1.0752 kgm/s

The impulse given to the ball by the floor is 1.0752 kgm/s

                             

6 0
3 years ago
Observing and experimenting are two ways that ecologists can answer scientific questions. What are some differences between thes
Sav [38]

Answer:OBSERVING IS WHATCHING A OBJECT VERY CLOSELY AND EXPIERIMENTING IS WER YOUTEST ON A CERTAIN THING OR CREATURE OR MASS OR ELEMENT

Explanation:IM THE MYSTERY MAN WHOOSH

6 0
3 years ago
flat sheet is in the shape of a rectangle with sides of lengths 0.400 mm and 0.600 mm. The sheet is immersed in a uniform electr
Ksenya-84 [330]

Answer:

6.29591\times 10^{-6}\ N/C^2

Explanation:

Flux is given by

\phi=EAcos\theta

A = Area

A=0.4\times 10^{-3}\times 0.6\times 10^{-3}

E = Electric field = 76.7 N/C

Angle is given by

\theta=90-20\\\Rightarrow \theta=70^{\circ}

\phi=76.7\times 0.4\times 10^{-3}\times 0.6\times 10^{-3}\times cos70\\\Rightarrow \phi=6.29591\times 10^{-6}\ N/C^2

The flux through the sheet is 6.29591\times 10^{-6}\ N/C^2

6 0
3 years ago
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