All categories

3 years ago

Jump starting a car requires

Engineering

1 answer:

Paha777 [63]3 years ago

3 0

Answer:

To safely jump a start, follow these steps:

Take out your jumper cables. ...

Place both vehicles in Park or Neutral and shut off the ignition in both cars. ...

Attach one of the red clips to the positive terminal of your battery. ...

Attach the other red clip to the positive terminal of the other car.

You might be interested in

10.0 kmol of a 40.0 mol% methanol and 60.0 mol% water mixture is processed in a normal batch distillation system with a still po

Serggg [28]

Answer:

[a]. 0.49.

[2]. 0.536

[c]. 4.15 kmol; 5.84 kmol.

Explanation:

Without mincing words let's dive straight into the solution to the question above.

[a].

The initial external reflux ratio, LD that must be used = [(0.85 - 0.57)/ 0.85 - 0]/ [ 1 - (0.85 - 0.57)/ 0.85 - 0].

The initial external reflux ratio, LD that must be used = 0.329/ 1- 0.329 = 0.49.

[b].

The final external reflux ratio that must be used = [ 0.85 - 0.13/ 0.85 - 0]/ [ 1 - 0.85 - 0.13/ 0.85 - 0].

Hence, the final external reflux ratio that must be used =0.847/ 1 - 0.847 = 5.536.

[c].

The amount of distillate product that is withdrawn:

4 = 0.85 H(t) + 0.8 - 0.08.

H(t) = 4.15 kmol, and the value of Wfinal = 5.84 kmol.

3 0

3 years ago

Compute the first four central moments for the following data:

alina1380 [7]

Answer:

Compute the first four central moments for the following data:

i xi

1 45

2 22

3 53

4 84Explanation:

7 0

3 years ago

In the idealized Otto cycle, heat is added during: a. Isentropic Compression b. Constant (minimum) volume c. Constant (maximum)

docker41 [41]

Answer:

(b) Constant (minimum) volume

Explanation:

In the idealized Otto cycle there are 4 process that are

Reversible adiabatic compression
Addition of heat at constant volume
Reversible adiabatic expansion
Rejection of constant volume

So from above discussion we can see that heat is added when there is constant (minimum) volume which is given in option (b) so option (b) will be the correct answer

3 0

3 years ago

A reservoir manometer has vertical tubes of diameter D518 mm and d56 mm. The manometer liquid is Meriam red oil. Develop an alge

Nat2105 [25]

Answer:

Explanation:

Given that :

the diameter of the reservoir D = 18 mm

the diameter of the manometer d = 6 mm

For an equilibrium condition ; the pressure on both sides are said to be equal

∴

$\Delta \ P = \rho _{water} g \Delta h_{water} = \rho _{oil} g \Delta h_{oil}$

$\Delta \ P = \rho _{oil} g (x+L) ----- (1)$

According to conservation of volume:

$A*x = a*L$

$\dfrac{\pi}{4}D^2x = \dfrac{\pi}{4}d^2 L$

$x = ( \dfrac{d}{D})^2L$

Replacing x into (1) ; we have;

$\Delta \ P = \rho _{oil} g ( ( \dfrac{d}{D})^2L+L)$

$\Delta \ P = \rho _{oil} g \ L ( ( \dfrac{d}{D})^2+1)$

$L = \dfrac{\Delta \ P}{\rho _{oil} g \ ( ( \dfrac{d}{D})^2+1)}$

Thus; the liquid deflection is : $L = \dfrac{\Delta \ P}{\rho _{oil} g \ ( ( \dfrac{d}{D})^2+1)}$

when the applied pressure is equivalent to 25 mm of water (gage); the liquid deflection is:

$L = \dfrac{\Delta \ P}{\rho _{oil} g \ ( ( \dfrac{d}{D})^2+1)}$

$L = \dfrac{\rho_{water} \g \Delta \ h}{\rho _{water} SG_{oil}g \ ( ( \dfrac{d}{D})^2+1)}$

$L = \dfrac{\g \Delta \ h}{SG_{oil}g \ ( ( \dfrac{d}{D})^2+1)}$

$L = \dfrac{25}{0.827 ( ( \dfrac{6}{18})^2+1)}$

L = 27.21 mm

4 0

3 years ago

D 4.37. Assuming the availability of diodes for which vD = 0.75 V at iD = 1mA, design a circuit that utilizes four diodes connec

gregori [183]

Answer:

R = 0.5825 k ohm

Explanation:

given data

vD = 0.75 V

iD = 1mA

resistor R = 15-V

solution

we get here first vD that is

vD = $\frac{3.3}{4}$

vD = 0.825

iD = Ise × $e^{\frac{vD}{nvT}}$

n = 1

so we can say

$\frac{iD2}{iD1} = e^{\frac{vD2-vD1}{nvT}}$

so it will

iD2 = iD1 × $e^{\frac{vD2-vD1}{nvT}}$

put here value and we get

iD2 = 1 × $e^{\frac{0.825-0.75}{25}}$

iD2 = 1 × $e^{3}$

iD2 = 20.086 mA

R will be

R = $\frac{15-3.3}{ID}$

R = 0.5825 k ohm

6 0

4 years ago