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Artyom0805 [142]
3 years ago
8

Technician A says that an oxygen sensor can have​ one, two,​ three, four, or more​ wires, depending on the style and design. Tec

hnician B says that if the exhaust has little​ oxygen, the voltage of the oxygen sensor will be close to 1 volt​ (1000 mV) and close to zero if there is high oxygen content in the exhaust. Who is​ right?
Physics
1 answer:
alisha [4.7K]3 years ago
7 0

Answer:

Both technicians A and B are right.

Explanation:

-The most common application is to measure the exhaust-gas concentration of oxygen.The probe typically has four wires attached to it: two for the lambda output, and two. The number of wires depends on design and style.

-An oxygen sensor will typically generate up to about 0.9 volts(\approx 1.0V) when the fuel mixture is rich and there is little unburned oxygen in the exhaust. When the mixture is lean, the sensor's output voltage will drop down to about 0.1( \approx 0.0V)volts

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There is a 3-kg toy cart moving at 4 m/s. Calculate the kinetic<br> energy of the cart.
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The kinetic energy of the cart is 24 J.

<u>Explanation:</u>

The acceleration of a given mass from rest to the velocity is known as kinetic energy. It gains energy from acceleration and remains in this state until the speed of the object changes.  

The kinetic energy is the given by,

                           K.E = 1/2 mv^2

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An electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric
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Answer:

1.) 11 km/s

2.) 9.03 × 10^-5 metres

Explanation:

Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.

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(a) What is the speed of the electron 1.3 ns after entering this region?

E = F/q

F = Eq

Ma = Eq

M × V/t = Eq

Substitute all the parameters into the formula

9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19

V = 7.68×10^-18 /7.0×10^-22

V = 10971.43 m/s

V = 11 Km/s approximately

(b) How far does the electron travel during the 1.3 ns interval?

The initial velocity U = 64 km/s

S = ut + 1/2at^2

S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2

S =8.32×10^-5 + 7.13×10^-6

S = 9.03 × 10^-5 metres

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