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olganol [36]
3 years ago
12

What is the difference between a stepped and a non-stepped ECT circuit? ​

Engineering
1 answer:
galina1969 [7]3 years ago
5 0

Answer:

A stepped circuit is designed in the engine coolant temperature (ECT) sensor within a powertrain control module (PCM) to increase the sensor's accuracy. A simple non-stepped ECT sensor circuit has a specific resistance which increases or decreases according to the changes in engine coolant temperature.

Explanation:

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Vector A extends from the origin to a point having polar coordinates (7, 70ᵒ ) and vector B extends from the origin to a point h
yaroslaw [1]

Answer:

13.95

Explanation:

Given :

Vector A polar coordinates = ( 7, 70° )

Vector B polar coordinates = ( 4, 130° )

To find A . B we  will

A ( r , ∅ ) = ( 7, 70 )

A = rcos∅ + rsin∅

therefore ; A  = 2.394i + 6.57j

B ( r , ∅ ) = ( 4, 130° )

B = rcos∅ + rsin∅

therefore ;  B = -2.57i + 3.06j

Hence ; A .B

( 2.394 i + 6.57j ) . ( -2.57 + 3.06j ) = 13.95

8 0
2 years ago
The following electrical characteristics have been determined for both intrinsic and p-type extrinsic gallium antimonide (GaSb)
xxTIMURxx [149]

Answer:

0.5m^2/Vs and 0.14m^2/Vs

Explanation:

To calculate the mobility of electron and mobility of hole for gallium antimonide we have,

\sigma = n|e|\mu_e+p|e|\mu_h (S)

Where

e= charge of electron

n= number of electrons

p= number of holes

\mu_e= mobility of electron

\mu_h=mobility of holes

\sigma = electrical conductivity

Making the substitution in (S)

Mobility of electron

8.9*10^4=(8.7*10^{23}*(-1.602*10^{-19})*\mu_e)+(8.7*10^{23}*(-1.602*10^{-19})*\mu_h)

0.639=\mu_e+\mu_h

Mobility of hole in (S)

2.3*10^5 = (7.6*10^{22}*(-1.602*10^{-19})*\mu_e)+(1*10^{25}*(-1.602*10^{-19}*\mu_h))

0.1436 = 7.6*10^{-3}\mu_e+\mu_h

Then, solving the equation:

0.639=\mu_e+\mu_h (1)

0.1436 = 7.6*10^{-3}\mu_e+\mu_h (2)

We have,

Mobility of electron \mu_e = 0.5m^2/V.s

Mobility of hole is \mu_h = 0.14m^2/V.s

6 0
3 years ago
Cite the phases that are present and the phase compositions for the following alloys: (a) 15 wt% Sn - 85 wt% Pb at 100 o C. (b)
Novosadov [1.4K]

Answer:

a)  ∝  and β

   The phase compositions are :

    C_{\alpha } = 5wt% Sn - 95 wt% Pb

    C_{\beta } =  98 wt% Sn - 2wt% Pb

b)

The phase is; ∝  

The phase compositions is;   82 wt% Sn - 91.8 wt% Pb

Explanation:

a) 15 wt% Sn - 85 wt% Pb at 100⁰C.

The phases are ; ∝  and β

The phase compositions are :

C_{\alpha } = 5wt% Sn - 95 wt% Pb

C_{\beta } =  98 wt% Sn - 2wt% Pb

b) 1.25 kg of Sn and 14 kg Pb at 200⁰C

The phase is ; ∝  

The phase compositions is;  82 wt% Sn - 91.8 wt% Pb

Csn = 1.25 * 100 / 1.25 + 14 = 8.2 wt%

Cpb = 14 * 100 / 1.25 + 14 = 91.8 wt%

6 0
2 years ago
The ???? − i relationship for an electromagnetic system is given by ???? = 1.2i1/2 g where g is the air-gap length. For current
Artemon [7]

Answer:

a) The mechanical force is -226.2 N

b) Using the coenergy the mechanical force is -226.2 N

Explanation:

a) Energy of the system:

\lambda =\frac{1.2*i^{1/2} }{g} \\i=(\frac{\lambda g}{1.2} )^{2}

\frac{\delta w_{f} }{\delta g} =\frac{g^{2}\lambda ^{3}  }{3*1.2^{2} }

f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{g^{2}\lambda ^{3}  }{3*1.2^{2} }

If i = 2A and g = 10 cm

\lambda =\frac{1.2*i^{1/2} }{g} =\frac{1.2*2^{1/2} }{10x10^{-2} } =16.97

f_{m}=-\frac{g^{2}\lambda ^{3}  }{3*1.2^{2} }=-\frac{16.97^{3}*2*0.1 }{3*1.2^{2} } =-226.2N

b) Using the coenergy of the system:

f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{1.2*2*i^{3/2}  }{3*g^{2} }=-\frac{1.2*2*2^{3/2} }{3*0.1^{2} } =-226.2N

8 0
2 years ago
You are riding in an elevator that is going up at 10 ft/s. You are holding your cell phone 5 ft above the floor when it suddenly
Luba_88 [7]

Answer:

It falls at the same speed in both cases.

Explanation:

If I were standing still the phone would be in free fall after slipping out of my hand.

I set a frame of reference with origin on the ground and the positive Y axis pointing up.

It would slip at t0 = 0, from a position Y0 = 5 ft, with a speed of Vy0 = 0.

It would be subject to an gravitational acceleration of -32.2 ft/s^2.

Since acceleration is constant:

Y(t) = Y0 + Vy0 * t + 1/2 * 4 * t^2

When it hits the floor at t1 it will be at Y(t1) = 0

0 = 5 + 0 * t1 - 16.1 * t1^2

16.1 * t1^2 = 5

t1^2 = 5 / 16.1

t1 = \sqrt{0.31} = 0.55 s

If the elevator is standing still it would take 0.55 s to hit the ground.

Now, if the elevator is moving up at 10 ft/s.

The frame of reference will have its origin at the place the floor of the elevator is at t = 0, and stay there as the elevetor moves. The floor of trhe elevator will have a position of Ye = 10 * t

Vy0 = 10 ft/s because it will be moving initially at the same speed as the elevator.

And it will hit the floor of the elevator not at 0, but at

Ye = 10 * t2

So:

10 * t2 = 5 + 10 * t2 - 16.1 * t2^2

0 = 5 - 16.1 * t2^2

16.1 * t1^2 = 5

t1^2 = 5 / 16.1

t1 = \sqrt{0.31} = 0.55 s

It falls at the same speed in both cases.

4 0
3 years ago
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