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4 years ago

How do you remove a manual transmission?

Engineering

1 answer:

Citrus2011 [14]4 years ago

4 0

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:

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What is the function of the following: 1- Oil rings 2- Flywheel 3- Timing gears

Tanya [424]

Answer:

Explanation:

The functions of the following are as follows:

1). Oil Rings:

Regulates oil within cylinder walls.
Helps to keep cylinder walls lubricated
Prevent Heat transfer

Reduce friction between piston and cylinder

2). Fly Wheel:

It is used to store rotational energy
It resist any change in rotational speed due to its moment of inertia
It helps to control the orientation of the mechanical system

3). Timing gears:

It provides synchronization in the rotation of crankshaft and the camshaft so as to provide proper valve opening and closing time during each cylinder's stroke(intake and exhaust).

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3 years ago

I have a Dutch oven that looks like this what do I do?

dangina [55]

Answer:

baking soda and vinegar dish soap

Explanation:

it will create a bubbles and let it sit for 3 hours and it will go away

3 0

3 years ago

This elementary problem begins to explore propagation delayand transmission delay, two central concepts in data networking. Cons

telo118 [61]

Explanation:

(a)

Here, distance between hosts A and B is m meters and, propagation speed along the link is s meter/sec

Hence, propagation delay, $d_{prop} = m/sec (s)$

(b)

Here, size of the packet is L bits

And the transmission rate of the link is R bps

Hence, the transmission time of the packet, $d_{trans} = L/R$

(c)

As we know, end-to-end delay or total no delay,

$\mathrm{d}_{\text {nodal }}=\mathrm{d}_{\text {proc }}+\mathrm{d}_{\text {quar }}+d_{\max }+d_{\text {prop }}$

$Here, $\mathrm{d}_{\text {rroc }}$ and $\mathrm{d}_{\text {quat }}$ \\Hence, $\mathrm{d}_{\text {rodal }}=\mathrm{d}_{\text {trass }}+\mathrm{d}_{\text {prop }}$ \\We know, $\mathrm{d}_{\text {trax }}=\mathrm{L} / \mathrm{R}$ sec and $\mathrm{d}_{\text {vapp }}=\mathrm{m} / \mathrm{s}$ sec$ $\text { Hence, } {d_{\text {nodal }}}=\mathrm{L} / \mathrm{R}+\mathrm{m} / \mathrm{s} \text { seconds }$

(d)

The expression, time $time $t=d_{\text {trans }}$$ means the\at time since transmission started is equal to transmission delay.

As we know, transmission delay is the time taken by host to push out the packet.

Hence, at $time $t=d_{\text {trans }}$$ the last bit of the packet has been pushed out or transmitted.

(e)

If $\ d_{prop} >d_{trans}$

Then, at $time $t=d_{\text {trans }}$$ the bit has been transmitted from host A, but to condition (1), the first bit has not reached B.

(f)

If $\ d_{prop}$

Then, at $time $t=d_{\text {trans }}$$ , the first bit has reached destination on B

Here, $s=2.5 \times 10^{8} \mathrm{sec}$

$\begin{aligned}&\mathrm{L}=100 \mathrm{Bits} \text { and }\\&\mathrm{R}=28 \mathrm{kbps} \text { or } 28 \times 1000 \mathrm{bps}\end{aligned}$

It's given that $\ d_{prop} =d_{trans}$

Hence,

$\begin{aligned}\ & \frac{L}{R}=\frac{m}{s} \\m &=s \frac{L}{R} \\&=\frac{2.5 \times 10^{8} \times 100}{28 \times 1000} \\&=892.9 \mathrm{km}\end{aligned}$

5 0

3 years ago

Solar ovens use heat from the Sun to cook food. They work on a principle similar to that of solar panels used to heat homes. Wha

marishachu [46]

Explanation:

Absorption of the sun's energy (heat) in solar cooking is best achieved when a surface is dark in color, thus the most common solar oven interiors are usually black in color as well as the color of the cookware used for cooking the food. Dark colors absorb the heat, whereas light colors do not absorb heat well.

Metal reflectors should be positioned around the oven to maximize light input to be the most effective.

4 0

3 years ago

Read 2 more answers

The driving force for fluid flow is the pressure difference, and a pump operates by raising the pressure of a fluid (by converti

Lady_Fox [76]

Answer:

The maximum possible volume flow of gasoline is 0.543 m^3/s

Explanation:

Power = pressure differential × volume flow rate

Power = 3.8 kW

Pressure differential = 7 kPa

Volume flow rate = power ÷ pressure differential = 3.8 ÷ 7 = 0.543 m^3/s

3 0

4 years ago