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Alona [7]
3 years ago
7

Technician A says that a MAF sensor is a high-authority sensor and is responsible for determining the fuel needs of the engine b

ased on the measured amount of air entering the engine. Technician B says that a cold wire MAF sensor uses the electronics in the sensor itself to heat a wire 20°C below the temperature of the air entering the engine. Who is right?
A) Technician A
B) Technician B
C) Both technicians
D) Neither technician
Physics
1 answer:
xz_007 [3.2K]3 years ago
8 0

Answer:

a. Technician A

Explanation:

Technician A says that a MAF sensor is a high-authority sensor and is responsible for determining the fuel needs of the engine based on the measured amount of air entering the engine. Technician B says that a cold wire MAF sensor uses the electronics in the sensor itself to heat a wire 20°C below the temperature of the air entering the engine. Who is right

MAF wich stands for  mass airflow sensor determines the mass of air flowing into the engine's air intake system. ... , the wire cools When air flows past the wire, decreasing its resistance, thereby more current flows through the circuit. When the MAf goes bad, it can not sense the amount of air intake into the engine.

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What are the damages or effects of sewage dumps into natural water sources???
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Greg is giving a slide show presentation to a large audience. How might a laser best help him with the presentation?
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A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00
mr_godi [17]

Answer:

Explanation:

initial velocity, u = 8 m/s

vertical height, h = 1 m

θ = 40°

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h = ut + 1/2 at²

1 = 8 Sin 40 x t + 0.5 x 9.8 t²

1 = 5.14 t - 4.9t²

4.9t² - 5.14 t + 1 = 0

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5 0
3 years ago
Block A of mass M is at rest and attached to the top of a spring. The block compresses the spring a distance d from its uncompre
Anni [7]

Answer:

a)  k = Mg / d , b)   v = √2gh , c)  v_{f} = \frac{2}{3} \ \sqrt{2gh},  d)   x² + 6d x - \frac{8}{3} dh = 0

e)the spring must compress a greater distance.

Explanation:

a) when the block of mass M is placed on the spring, we have an equilibrium condition,

             ∑ F  = 0

             F_{e}- W = 0

             k d = Mg

             k = Mg / d

b) let's use the concepts of energy to find the velocity of the block just before the collision

starting point. Position when released

          Em₀ = U = m g h

lowest point. Right at the point of shock

          Em_{f} = K = ½ m v²2

as there is no friction, energy is conserved

          Em₀ = Em_{f}

          mg h = ½ m v²

          v = √2gh

         

c) The velocity of the two blocks after the collision, we define a system formed by the two blocks, in such a way that the forces during the collision are internal and the moment is conserved

initial instant. Just before the crash

          p₀ = 2M v + M 0

final instant. Just after the shock, before the spring compression begins

         p_{f} = (2M + M) v_{f}

 the moment is preserved

          p₀ = p_{f}

          2M v = 3M v_{f}

          v_{f} = ⅔ v

          v_{f} = \frac{2}{3} \ \sqrt{2gh}

d) now we work with the joined system after the collision, let's use the concepts of energy

starting point. After shock, before beginning spring compression

        Em₀ = K = ½ (3M) v_{f}^2

        Em₀ = 3/2 M (\frac{2}{3} \ \sqrt{2gh})²

        Em₀ = 4/3 M gh

final point. With the spring fully compressed

       Em_f = K_e + U = ½ k x² + (3M) g x

in this case we have taken the zero of gravitational potential energy at the point where the blocks collide, as there is no friction, the energy is conserved

         Em₀ = Em_f

        4/3 M g h = ½ k x² + 3M g x

        ½ k x² + 3Mg x - 4/3 Mgh = 0

we substitute the expression for k

         \frac{1}{2} (\frac{Mg}{d}) x² + 3Mg x - \frac{4}{3} Mgh = 0

          \frac{x^{2} }{2d} + 3 x - \frac{4}{3}h = 0

to find the value of the spring compression, the second degree equation must be solved

          x² + 6d x - \frac{8}{3} dh = 0

         x = [-6d ±\sqrt{(36 d^{2} - 4 \frac{8}{3} dh)  } ] / 2

         x = [-6d ± 6d \sqrt{ 1 -  \frac{32}{3 \ 36}  \ \frac{h}{d}    }  ]/2

         x = 3d ( -1±  \sqrt{ 1 - 0.296 \frac{h}{d}   }  )

e) If the collision elastic force would not lose any part of the kinetic energy during the collision, therefore the speed of the block of mass M would be much higher and therefore the spring must compress a greater distance.

8 0
3 years ago
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