Answer:
Step 1
Elevate the front of the vehicle using the floor jack and support the vehicle with two jack stands. Make sure the vehicle is stable.
Step 2
Disengage all electrical components connected to the transmission. Indicate by marking the position of the drive shaft for its reinstallation. From the output shaft, remove the rear U joint. Jam the cloth to keep the liquid from dripping out of the extension housing.
Step 3
Loosen the shift linkages and the speedometer cable from the transmission manually. Place the transmission jack under the transmission, and then take a socket wrench and remove the support nut, the cross-member, and the rear support insulator from the rear engine. Support the engine with a jack stand and use the transmission jack to withdraw the transmission toward the rear of the vehicle.
Explanation:
Answer:
b. spark plugs
Explanation:
Diesel engines are characterized in that the mechanism that activates the explosion of fuel is high pressures, so when the piston reaches the top of the cylinder, the pressure of the air-diessel mixture is so high that it causes the explosion, this It is what generates the power in a diesel engine.
Answer:
248.756 mV
49.7265 µA
Explanation:
The Thevenin equivalent source at one terminal of the bridge is ...
voltage: (100 V)(1000/(1000 +1000) = 50 V
impedance: 1000 || 1000 = (1000)(1000)/(1000 +1000) = 500 Ω
The Thevenin equivalent source at the other terminal of the bridge is ...
voltage = (100 V)(1010/(1000 +1010) = 100(101/201) ≈ 50 50/201 V
impedance: 1000 || 1010 = (1000)(1010)/(1000 +1010) = 502 98/201 Ω
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The open-circuit voltage is the difference between these terminal voltages:
(50 50/201) -(50) = 50/201 V ≈ 0.248756 V . . . . open-circuit voltage
__
The current that would flow is given by the open-circuit voltage divided by the sum of the source resistance and the load resistance:
(50/201 V)/(500 +502 98/201 +4000) = 1/20110 A ≈ 49.7265 µA
Answer:112.376 s
Explanation:
Given
Using Newton's law of cooling
=
=
7=
Taking log both side
t=112.376sec
Answer:
367.43 mm²
Explanation:
Given:
Flow rate, Q = 0.7 L/s
1000 L = 1 m³ = 10⁹ mm³
thus,
1 L = 10⁶ mm³
Therefore,
Q = 0.7 × 10⁶ mm³/s
Cross-sectional area at the top of the sprue = 750 mm²
Length of the sprue = 185 mm
Now,
Velocity = 
where, g is the acceleration due to gravity = 9.81 m/s²
h is the height through which flow is taking place = 185 mm = 0.185
thus,
Velocity = 
or
velocity = 1.9051 m/s = 1905.1 mm/s
Also,
Q = Area × Velocity
thus,
0.7 × 10⁶ = Area × 1905.1
or
Area = 367.43 mm²
