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2 years ago

Replace oxygen sensor but light still on

Physics

1 answer:

otez555 [7]2 years ago

8 0

The O2 sensor does produce its own voltage,but there are also two wires.

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The speed of light in a material is found to be 2.07 x 108 m/s. What is the most likely material from the options below if the s

Paraphin [41]

Answer:

1.3636

Explanation:

Write the expression for the refractive index.

n=c/v

c= speed of light in air

v= speed of light in material

=(3×10^8 m/s)/(2.2×10^8 m/s)

=1.3636

4 0

3 years ago

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A bullet with an initial kinetic energy of 400 J strikes a wooden block where a 8000 N resistive force stops the bullet. What is

Sindrei [870]

Answer:

d = 0.05 [m] = 50 [mm]

Explanation:

We must remember the principle of conservation of energy which tells us that energy is transformed from one way to another. For this case, the initial kinetic energy is transformed into useful work that is equal to the product of force by distance.

$E_{k}=F*d\\400 = 8000*d\\d = 0.05 [m] = 50 [mm]$

5 0

2 years ago

If a bar of copper is brought near a magnet, the copper bar will be

Alecsey [184]

It will be unaffected by the magnet because it has no magnetic field. If you were to maybe have electricity going through it is the only way it would have anything to do with the magnet.

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3 years ago

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A piston-cylinder device contains Helium gas initially at 150 kPa, 20 o C, and 0.5m 3 . The helium is now compressed in a polytr

Molodets [167]

Answer:

Explanation:

Given

$P_1=150 kPa$

$T_1=20^{\circ}C$

$V_1=0.5 m^3$

$T_2=140^{\circ}C$

$P_2=400 kPa$

R for Helium $R=2.076$

$c_v=3.115 kJ/kg-K$

mass of gas $m=\frac{P_1V_1}{RT_1}$

$m=\frac{150\times 0.5}{2.076\times 293}$

$m=0.123 kg$

Similarly $V_2$ can be found

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$V_2=0.264 m^3$

Work done $W=\int_{V_1}^{V_2}PdV$

$W=\frac{P_2V_2-P_1V_1}{n-1}$

$W=\frac{mR(T_2_T_1)}{n-1}$

Since it is a polytropic Process

therefore $PV^n=c$

$P_1V_1^n=P_2V_2^n$

$(\frac{V_1}{V_2})^n=\frac{P_2}{P_1}$

$(\frac{0.5}{0.264})^n=\frac{400}{150}$

$n=\frac{\ln 2.66}{\ln 1.893}$

$n=1.533$

$W=\frac{0.123\times 2.076(140-20)}{1.533-1}$

$W=57.48 kJ$

From Energy balance

$E_{in}-E_{out}=\Delta E_{system}$

Neglecting kinetic and Potential Energy change

$Q_{in}+W_{in}=change\ in\ Internal\ Energy$

Change in Internal Energy $\Delta U=u_2-u_1$

$\Delta U=mc_v(T_2-T_1)$

$\Delta U=0.123\times 3.115(140-20)$

$\Delta U=45.977 kJ$

$Q_{in}+57.48=45.977$

$Q_{in}=-11.50 kJ$

i.e. Heat is being removed

3 0

3 years ago

How do atoms of copper differ from atoms of aluminum? *

Degger [83]

Answer:

The copper atoms are heavier than the aluminum atoms. The copper atoms are smaller than the aluminum atoms so more copper atoms fit in the same volume. 2. Copper is more dense than aluminum.

HOPE THIS HELPS !!!!!

8 0

2 years ago